LIBRARY 

OF  THE 

University  of  California. 

Mrs.  SARAH  P.  WALSWORTH. 

Received  October,  1894. 
^Accessions  No .  S^/Z/C?   .      Class  No. 


/ 


Digitized  by  the  Internet  Archive 

in  2008  with  funding  from 

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http://www.archive.org/details/additionalbourdOOdavirich 


KEY 


DAVIES'  BOURDON, 


MANY  ADDITIONAL    EXAMPLES,  ILLUSTRATING   THE 
ALGEBRAIC  ANALYSIS: 


ALSO, 


A  SOLUTION  OF  ALL  THE  DIFFICULT  EXAMPLES 
IN  DAVIES'  LEGENDRE. 


UWVBRSIT 


OJf 


4km% 


NEW    YORK 
A.  S.  BARNES  &  Co.,  Ill  &  113  WILLIAM  STREET, 

(CORNER   OF  JOHN   STREET.) 
BOLD   BY  BOOKSELLERS,  GENERALLY,  THROUGHOUT  THE  UNITED   STATES. 

0  1866. 


ID  33 


(f^s\V^ 


Entered  according  to  act  of  Congress,  in  the  year  eighteen  hundred  and  fifty-six^ 

BY    CHARLES    DAVIES, 

In  the  Clerk's  Office  of  the  District  Court  of  the  United  States,  for  the  Southern 

District  of  New  York. 


JONES      &      DE  N YSE , 
8TEREOTYPERS    AND    ELECTROTYPERS, 
183  William-Street. 


0f  THK 

ViiviasxTY; 

PREFACE. 


A  wide  difference  of  opinion  is  known  to  exist  among  teachers 
in  regard  to  the  value  of  a  Key  to  any  mathematical  work,  and 
it  is  perhaps  yet  undecided  whether  a  Key  is  a  help  or  a 
hindrance. 

If  a  Key  is  designed  to  supersede  the  necessity  of  investiga- 
tion and  labor  on  the  part  of  the  teacher;  to  present  to  his 
mind  every  combination  of  thought  which  ought  to  be  suggested 
by  a  problem,  and  to  permit  him  to  float  sluggishly  along  the 
current  of  ideas  developed  by  the  author,  it  would  certainly  do 
great  harm,  and  should  be  excluded  from  every  school. 

If,  on  the  contrary,  a  Key  is  so  constructed  as  to  suggest 
ideas,  both  in  regard  to  particular  questions  and  general  science, 
which  the  Text-book  might  not  impart;  if  it  develops  methods 
of  solution  too  particular  or  too  elaborate  to  find  a  place  in  the 
text ;  if  it  is  mainly  designed  to  lessen  the  mechanical  labor  of 
teaching,  rather  than  the  labor  of  study  and  investigation ;  it 
may,  in  the  hands  of  a  good  teacher,  prove  a  valuable  auxiliary. 

The    Key   to  Bourdon   is  intended    to  answer,    precisely,    this 


IV  PREFACE. 

end.  The  principles  developed  in  the  text  are  explained  and 
illustrated  by  means  of  numerous  examples,  and  these  are  all 
wrought  in  the  Key  by  methods  which  accord  with  and  make 
evident  the  principles  themselves.  The  Key,  therefore,  not  only 
explains  the  various  questions,  but  is  a  commentary  on  the  text 
itself. 

Nothing  is  more  gratifying  to  an  ambitious  teacher  than  to 
push  forward  the  investigations  of  his  pupils  beyond  the  limits 
of  the  text  book.  To  aid  him  in  an  undertaking  so  useful  to 
himself  and  to  them,  an  Appendix  has  been  added,  containing  a 
copious  collection  of  Practical  Examples.  Many  of  the  solutions 
are  quite  curious  and  instructive ;  and  taken  in  connection  with 
those  embraced  in  the  Text,  form  a  full  and  complete  system  of 
Algebraic  Analysis. 

The  Problems  comprising  the  "Application  of  Algebra  to 
Geometry,"  at  the  end  of  Legendre,  are,  many  of  them,  quite 
difficult  of  solution. 

The  many  letters  which  I  have  received  from  Teachers  and 
Pupils,  in  regard  to  the  best  solutions  of  these  questions,  have 
suggested  the  desirableness  of  furnishing,  in  the  present  work 
those  which  have  been  most  approved.  They  are  a  collection  of 
problems  that  have  been  often  solved,  and  the  solutions  may  be 
studied  with  great  profit  by  every  one  seeking  mathematical 
knowledge. 

Fishkill  Landing,  ) 

July,  1856.        J  \ 


V 


Ol 


INTRODUCTION 


ALGEBRA. 


1.  Otf  an  analysis  of  the  subject  of  Algebra,  we 
think  it  will  appear  that  the  subject  itself  presents  no 
serious  difficulties,  and  that  most  of  the  embarrassment 
which  is  experienced  by  the  pupil  in  gaining  a  knowl- 
edge of  its  principles,  as  well  as  in  their  applications, 
arises  from  not  attending  sufficiently  to  the  language 
or  signs  of  the  thoughts  which  are  combined  in  the 
reasonings.  At  the  hazard,  therefore,  of  being  a  little 
diffuse,  I  shall  begin  with  the  very  elements  of  the 
algebraic  language,  and  explain,  with  much  minute- 
ness, the  exact  signification  of  the  characters  that  stand 
for  the  quantities  which  are  the  subjects  of  the  analy- 
sis ;  and  also  of  those  signs  which  indicate  the  several 
operations  to  be  performed  on  the  quantities. 


Algebra. 
Difficulties 

How  over- 
come. 

Language. 


Characters 
which  repre- 
sent quantity. 

Sign«. 


2.  The  quantities  which  are  the  subjects  of  the  ^uantit>'es- 
algebraic  analysis  may  be  divided  into  two  classes :  How  divided 
those  which  are  known  or  given,  and  those  which  are 
unknown  or  sought.  The  known  are  uniformily  repre- 
sented by  the  first  letters  of  the  alphabet,  ez,  b,  c,  d9 
&c. ;  and  the  unknown  by  the  final  letters,  x,  y,  z, 
V,   w.   &c. 


How  repre- 
sented. 


6 


Introduction. 


May  be  in- 
creased or 
diminished. 

Five  opera- 
tions. 


First 

Second. 
Third. 

Fourth. 
Fifth. 

Exception. 


3.  Quantity  is  susceptible  of  being  increased  or 
diminished ;  and  there  are  five  operations  which  ean 
be  performed  upon  a  quantity  that  will  give  results 
differing  from  the  quantity  itself,  viz.  : 

1st.  To  add  it  to  itself  or  to  some  other  quantity ; 

2d.    To  subtract  some  other  quantity  from  it ; 

3d.    To  multiply  it  by  a  number; 

4th.  To  divide  it  ; 

5th.  To  extract  a  root  of  it. 

The  cases  in  which  the  multiplier  or  divisor  is  1, 
are  of  course  excepted;  as  also  the  case  where  a  root 
is  to  be  extracted  of  1. 


Signs.  4.   The   five   signs    which    denote    these   operations 

Elements  are  too  well  known  to  be  repeated  here.     These,  with 

of  the 

Algebraic  the  signs  of  equality  and  inequality,  the  letters  of  the 

anguage.  a]p}iaketi  anci  tne  figures  which  are  employed,  make  up 

its  words  and  the    elements  of   the  algebraic    language.     The    words 

p  rases  ^^    phrases    of    the    algebraic,    like    those    of    every 

How  inter-  other   language,  are    to    be    taken    in    connection  with 

pTete  "  each  other,  and  are  not    to  be   interpreted  as  separate 

.  and  isolated  symbols. 


Bymbols  of 
quantity. 


5.  The  symbols  of  quantity  are  designed  to  repre- 
sent quantity  in  general,  whether  abstract  or  concrete, 
whether  known  or  unknown ;  and  the  signs  which  in- 
dicate the  operates  to  be  performed  on  the  quanti- 
ties are  to  be  interpreted  in  a  sense  equally  general. 
When  the  sign  plus  is  written,  it  indicates  that  the 
Signs  pins  an  a  quantity  before    which    it   is  placed  is  to  be  added  to 

minus. 

some  other    quantity  :  and   the  sign  minus   implies  the 


General. 


Examples. 


INTRODUCTION.  7 

existence  of  a  minuend,  from  which  the  subtrahend  is 

to  be  taken.     One  thing  should  be  observed  in  regard  Signs  have  n« 

effect  on  the 

to  the  signs  which  indicate  the  operations  that  are  to       nature  of 

be   performed   on  quantities,  viz. :    they  do   not   at   all     a  quan  l  r' 

affect  or  change   the   nature  of  the  quantity  before   or 

after  which   they  are  written,  but  merely  indicate  what 

is  to  be  done  with   the  quantity.     In  Algebra,  for  ex-     Examples: 

In  Algebra. 

ample,  the  minus  sign  merely  indicates  that  the  quan- 
tity before  which  it  is  written  is  to  be  subtracted  from 
some  other  quantity ;  and  in  Analytical  Geometry,  that    in  Analytical 

Geometry. 

the  line  before  which  it  falls  is  estimated  in  a  contrary 
direction  from  that  in  which  it  would  have  been  reck- 
oned, had  it  had  the  sign  plus ;  but  in  neither  case  is 
the  nature  of  the  quantity  itself  different  from  what 
it  would  have  been  had  the  sign  been  plus. 

The   interpretation   of   the    language    of  Algebra   is  interpretation 

of  the 

the  first  thing  to  which  the  attention  of  a  pupil  should      language : 

be  directed ;  and  he  should  be  drilled  on  the  meaning 

and   import   of  the   symbols,   until    their    significations 

and  uses  are  as  familiar  as  the  sounds  and  combina-    Its  neces81ty« 

tions  of  the  letters  of  the  alphabet. 

6.   Beginning   with    the  elements  of  the   language,      Element* 

explained 

let  any  number  or  quantity  be  designated  by  the  letter 
a,  and  let  it  be  required  to  add  this  letter  to  itself 
and   find   the   result   or   sum.      The   addition   will    be 

expressed  by 

a  +  a  =  the  sum. 

But   how   is  the   sum   to   be   expressed?     By   simply    Signification, 
regarding  a  as  one  a,  or  la,  and  then  observing  that 
one  a  and  one  a,  make  two  a's    or  2a :  hence, 


O  INTRODUCTION. 

a  -\-  a  =  2a; 

and  thus  we  place  a  figure  before  a  letter  to  indicate 
how  many  times  it  is  taken.     Such  figure  is  called  a 
Co-efEciont.     Co-efficient. 

Preduct:  7.    The   product   of    several    numbers   is   indicated 

by  the  sign  of  multiplication,  or  by  simply  writing  the 
letters  which  represent  the  numbers  by  the  side  of 
each  other.      Thus, 

how  indicated.  axbxcxdxf,     or     abcdf, 

indicates  the .  continued  product  of %  a,  b,  c,  d,  and  /, 
Factor.  and  each  letter  is  called  a  factor  of  the  product : 
hence,  a  factor  of  a  product  is  one  of  the  multipliers 
which  produce  it.  Any  figure,  as  5,  written  before  a 
product,  as 

5abcdf, 

Co-efficient  of   is  the  co-efficient  of  the  product,  and  shows  that  the 
a  pro  uc .      pr0(juct  is  taken  5  times. 

fiiuai  factors :         g^    jf   the   numbers  represented   by  a,  b,  c,  d,  and 
what  the      f9   were   equal    to   each    other,   they   would    each    be 

product 

becomes.      represented    by   a    single    letter   a,   and    the    product 
would   then  become 

How  K 

expresses  aXaXaXdXa  =  a-; 

that  is,  we  indicate  the  product  of  several  equal  fac- 
tors by  simply  writing  the  letter  once  and  placing  a 
figure  above  and  a  little  at  the  right  of  it,  to  indicate 


INTRODUCTION. 


9 


how  many  times  it  is  taken  as  a  factor.     The  figure     Exponent: 
so  written  is  called  an  exponent.     Hence,  an  exponent  where  written. 
is  a  simple  form  of    language    to  point  out  how  many 
equal  factors  are  employed. 


9.  The  division  of  one  quantity  by  another  is  indi- 
cated by  simply  writing  the  divisor  below  the  dividend, 
after  the  manner  of  a  fraction  ;  by  placing  it  on  the 
right  of  the  dividend  with  a  horizontal  line  and  two 
dots  between  them  ;  or  by  placing  it  on  the  right  with 
a  vertical  line  between  them :  thus  either  form  of 
expression  : 

-— >  b  -f-  a,     or     b\a, 


Division : 

how 
expressed 


Three  forms. 


indicates  the  division  of  b  by  a. 

10.    The   extraction  of  a   root   is   indicated  by  the        Roots : 

sign  y\     This  sign,  when  used  by  itself  indicates  the  how  indicated 
lowest  root,  viz.,  the  square  root.      If  any  other  root 
is  to  be  extracted,  as  the  third,  fourth,  fifth,  &c,  the        index; 

figure  marking  the  degree  of  the  root  is  written  above  where  written 
and  at  the  left  of  the  sign  ;  as, 

y    cube  root,  y    fourth  root,  &c. 


The  figure  so  written,  is  called  the  Index  of  the  root.    Language  far 
We  have   thus   given    the  very  simple  and   general 


operations 


language  by  which  we  indicate  each  of  the  five 
operations  that  may  be  performed  on  an  algebraic 
quantity,  and  every  process  in  Algebra  involves  one  or 
other  of  these  operations. 


OF  THE 


tTHIVBRSITTi 


10  INTRODUCTION. 

MINUS     SIGN. 

languid  U-    The   algebraic   symbols    are    divided    into   two 

classes    entirely    distinct    from    each    other — viz.,    the 

ko-w  divided,    letters  that  are  used  to  designate  the  quantities  which 

are  the   subjects   of  the   science,  and   the  signs  which 

are  employed  to  indicate  certain  operations  to  be  per- 

Aigcbraic       formed   on    those  quantities.      We   have    seen    that   all 

processes  : 

the  algebraic  processes  are  comprised  under  addition, 
ihei:  number,   subtraction,   multiplication,  division,  and  the  extraction 
Do-not  change   of  roots  ;    and  it   is   plain,  that  the  nature  of  a  quan- 
go6 "antitiL    ^J  *s   not  at  a^  cnangecl  by  prefixing  to  it  the  sign 
which   indicates  either  of  these  operations.     The  quan- 
tity denoted  by  the  letter  «,  for  example,  is  the  same, 
in  ever])  respect^  whatever  sign  may  be  prefixed  to  it ; 
that  is,  whether  it  be  added  to  another  quantity,  sub- 
tracted from  it,  whether  multiplied  or  divided  by  any 
number,  or  whether  we  extract  the  square  or  cube  or 
Algebraic       any  other  root  of  it.     The  algebraic  signs,  therefore, 
signs:        must   be   regarded    merely  as    indicating   operations   to 

how  regarded.  *  ° 

be    performed    on    quantity,    and   not   as    affecting   the 

nature  of  the  quantities  to  which  they  may  be  prefixed. 

Plus  and       \\re   say,  indeed,  that  quantities    are    plus   and   minus, 

Minus. 

but   this   is    an    abbreviated   language   to  express    that 
they  are  to  be  added  or  subtracted. 

Principles  of  jo    In   Algebra,  as   in  Arithmetic   and    Geometry 

the  science-  x    '  °  '  ' 

From  what  all  the  principles  of  the  science  are  deduced  from  tn< 
definitions  and  axioms  ;  and  the  rules  for  performing 
the  operations  are  but  directions  framed  in  conformity 

Example.  to  such  principles.  Having,  for  example,  fixed  b; 
definition,  the  power  of  the  minus  sign,  viz.,  that  an; 


INTRODUCTION.  11 

quantity  before  which  it  is  written,  shall  be  regarded 

as  to  be  subtracted  from  another  quantity,-  we  wish  to  What  *'e  wiah 

to  discover 

discover  the  process  of  performing  that  subtraction,  so 
as  to  deduce  therefrom  a  general  formula,  from  which 
we  can  frame  a  rule  applicable  to  all  similar  cases. 


SUBTRACTION. 

13.    Let   it   be   required,   for    example,  to  subtract    Subtraction. 

h 


a  —  c 


Process. 


Difference. 


from  b  the  difference  between  a  and  c. 
Now,  having  written  the  letters,  with 
their  proper  signs,  the  language  of  Al- 
gebra expresses  that  it  is  the  difference  only  between 
a  and  c,  which  is  to  be  taken  from  b ;  and  if  this  dif- 
ference were  known,  we  could  make  the  subtraction  at 
once.  But  the  nature  and  generality  of  the  algebraic 
symbols,  enable   us   to   indicute  operations,  merely,  and     Operations 

indicated. 

we  cannot  in  general  make  reductions  until  we  come 
to  the  final  result.  In  what  general  way,  therefore, 
can  we  indicate  the  true  difference  ? 


b  -a 


b  —  a  -f-  c 


Final  formula, 


If  we  indicate  the  subtraction  of  a 
from  b,  we  have  b  —  a  ;  but  then  we 
have  taken  away  too  much  from  b  by 
the  number  of  units  in  r;for  it  was  not  a,  but  the  dif- 
ference between  a  and  c  that  was  to  be  subtracted 
from  b.  Having  taken '  away  too  much,  the  remainder 
is  too  small  by  c :  hence,  if  c  be  added,  the  true  re- 
mainder will  be  expressed  by  b  —  a  -j-  c. 

Now,  by  analyzing  this  result,  we  see  that  the  sign     Analysis  of 
of  every  term  of  the   subtrahend    has    been   changed ; 
and  what  has  been  shown  with  respect  to  these  quan- 

OF  THI 


the  result. 


WVBRsitt; 


12 


INTRODUCTION, 


Generaiiza-  titles  is  equally  true  of  all  others  standing  in  the  same 
relation :  hence,  we  have  the  following  general  rule 
for  the  subtraction  of  algebraic  quantities  : 

Change   the  sign  of  every  term   of  the  subtrahend,  or 
Rule.         conceive  it  to  be  changed,  and  then  unite  the  quantities 
as  in  addition. 


Multiplica- 
tion. 


Signification 

of  the 
language. 


Process. 


a  -b 
c 


Its  nature. 


Principle  for 
the  signs. 


ac  —  be 


MULTIPLICATION. 

14.  Let  us  now  consider  the  case  of  multiplication, 
and  let  it  be  required  to  multiply  a  —  b  by  c.  The 
algebraic  language  expresses  that  the 
difference  between  a  and  b  is  to  be 
taken  as  many  times  as  there  are 
units  in  c.  If  we  knew  this  differ- 
ence, we  could  at  once  perform  the  multiplication. 
But  by  what  general  process  is  it  to  be  performed 
without  finding  that  difference  ?  If  we  take  a,  c  times, 
the  product  will  be  ac;  but  as  it  was  only  the  differ- 
ence between  a  and  b>  that  was  to  be  multiplied  by  c, 
this  product  ac  will  be  too  great  by  b  taken  c  times ; 
that  is,  the  true  product  will  be  expressed  by  ac  —  be: 
hence,  we  see,  that, 

If  a  quantity  having  a  plus  sign  be  multiplied  by 
another  quantity  having  also  a  plus  sign,  the  sign  of 
the  product  will  be  plus  ;  and  if  a  quantity  having  a 
minus  sign  be  multiplied  by  a  quantity  having  a  plus 
sign,   the  sign   of  the  product  will  be  minus. 


General  case:         15.   Let  us  now  take  the  most   general  case,  viz., 
that  in  which  it  is  required  to  multipy  a  —  b  by  c  —  a\ 


INTRODUCTION. 


13 


a  —  b 
c  —  d 

ac  —  be 

-  ad  -f  bd 

ac  —  be  —  ad  -f  bd 


It*  form. 


First  step. 


Let  us  again  observe  that  the  algebraic  language 
denotes  that  a  —  b  is  to  be  ta- 
ken as  many  times  as  there 
are  units  in  c  —  d ;  and  if  these 
two  differences  were  known, 
their  product  would  at  once 
form   the   product   required. 

First :  let  us  take  a  —  b  as 
many  times  as  there  are  units  in  c  ;  this  product, 
from  what  has  already  been  shown,  is  equal  to  ac  —  be. 
But  since  the  multiplier  is  not  c,  but  c  —  t/,  it  follows 
that  this  prodtfet  is  too  large  by  a  —  b 'taken  d  times  ; 
that  is,  by  ad  —  bd :  hence,  the  first  product  dimin-  Second  step 
ished  by  this  last,  will  give  the  true  product.  But,  by 
the  rule  for  subtraction,  this  difference  is  found  by  How  token, 
changing  the  signs  of  the  subtrahend,  and  then  uniting 
all  the  terms  as  in  addition  :  hence,  the  true  product 
is  expressed  by  ac.  —  be  —  ad  -f-  bd. 

By  analyzing  this  result,  and  employing  an  abbre- 
viated language,  we  have  the  following  general  prin- 
ciple to  which  the  signs  conform  in  multiplication,  viz. : 


Analysis  of 
the  result. 


Plus  multiplied  by  plus  gives  plus  in  the  product ; 
plus  multiplied  by  minus  gives  minus ;  minus  mul- 
tiplied by  plus  gives  minus ;  and  minus  multiplied  by 
minus  gives  plus  in   the  product. 


General 
Principle. 


16.  The  remark  is  often  made  by  pupils  that  the 
above  reasoning  appears  very  satisfactory  so  long  as 
the  quantities  are  presented  under  the  above  form ; 
but  why   will   —  b   multiplied    by   - 


Remark. 


Particular 

give    plus   bd?         case. 


14 


INTRODUCTION. 


How  can  the  product  of  two  negative  quantities  stand* 
mg  alone  be  plus  ? 

Minus  sign.  In  the  first  place,  the  minus  sign  being  prefixed  to 
t  and  d,  shows   that  in   an  algebraic  sense  they  do  not 

t»  interpre-  otand  by  themselves,  but  are  connected  with  other  quan- 
tities ;  and  if  they  are  not  so  connected,  the  minus 
sign  makes  no  difference  ;  for,  it  in  no  case  affects  the 
quantity,  but  merely  points  out  a  connection  with  other 
quantities.  Besides,  the  product  determined  above, 
being   independent  of  any  particular   value   attributed 

Fmm  «f  the  to  tne  letters  o,  b,  c,  and  d,  must  be  of  such  a  form  as 
product :       tQ  ke  true  for  a|j  vaiues  .   anc[  hence  for  the  case  in 

must  be  true 

for  quantities   which  a  and  c  are  each  equal   to  zero.      Making  this 

of  any  value 

\    supposition,  the  product  reduces  to  the  form  of  4-  bd. 
Signs  in       The  rules  for  the  signs  in  division  are  readily  deduced 
from   the  definition  of  division,  and  the  principles  al- 
ready laid  down. 


Zero  and 
Infinity. 


Ideas  not 
abstruse. 


ZERO     AND      INFINITY. 

17.  The  terms  zero  and  infinity  have  given  rise  to 
much  discussion,  and  been  regarded  as  presenting  diffi- 
culties not  easily  removed.  It  may  not  be  easy  to 
frame  a  form  of  language  that  shall  convey  to  a  mind, 
but  little  versed  in  mathematical  science,  the  precise 
ideas  which  these  terms  are  designed  to  express  ;  but 
we  are  unwilling  to  suppose  that  the  ideas  themselves 
are  beyond  the  grasp  of  an  ordinary  intellect.  The 
terms  are  used  to  designate  the  two  limits  of  Space 
and  Number. 

18.  Assuming  any  two  points  in  space,  and  joining 


INTRODUCTION. 


15 


Illustration, 

showing  th« 

meaning  of 

the  term  Zero. 


Illustration, 

showing  the 

meaning  of 

the  term 

Infinity. 


them  by  a  straight  line,  the  distance  between  the  points 
will  be  truly  indicated  by  the  length  of  this  line,  and 
this  length  may  be  expressed  numerically  by  the  num- 
ber of  times  which  the  line  contains  a  known  unit.  If 
now,  the  points  are  made  to  approach  each  other,  the 
length  of  the  line  will  diminish  as  the  points  come 
nearer  and  nearer  together,  until  at  length,  when  the 
two  points  become  one,  the  length  of  the  line  will 
disappear,  having  attained  its  limit,  which  is  called 
lero.  If,  on  the  contrary,  the  points  recede  from  each 
other,  the  length  of  the  line  joining  them  will  con- 
tinually increase ;  but  so  long  as  the  length  of  the 
line  can  be  expressed  in  terms  of  a  known  unit  of 
measure,  it  is  not  infinite.  But,  if  we  suppose  the 
points  removed,  so  that  any  known  unit  of  measure 
would  occupy  no  appreciable  portion  of  the  line,  then 
the  length  of  the  line  is   said  to  be  Infinite. 

19.  Assuming  one  as  the  unit  of  number,  and  ad- 
mitting the  self-evident  truth  that  it  may  be  increased 
or  diminished,  we  shall  have  no  difficulty  in  under- 
standing the  import  of  the  terms  zero  and  infinity, 
as  applied  to  number.  For,  if  we  suppose  the  unit 
one  to  be  continually  diminished,  by  division  or  other- 
wise, the  fractional  units  thus  arising  will  be  less  and  illustration, 
less,  and  in  proportion  as  we  continue  the  divisions, 
they  will  continue  to  diminish.  Now,  the  limit  or 
boundary  to  which  these  very  small  fractions  approach, 
is  called  Zero,  or  nothing.  So  long  as  the  fractional 
number  forms  an  appreciable  part  of  one,  it  is  not 
zero,  but  a  finite  fraction ;  and  the  term  zero  is  only 


The  terms 
Zero  and  In- 
finity applied 
to  numbers. 


Zero: 


16 


INTKODUCTION. 


applicable  to  that  which  forms  no  appreciable  part  of 
the  standard. 

illustration.  If,  on  the  other  hand,  we  suppose  a  number  to  be 
continually  increased,  the  relation  of  this  number  to  the 
unit  will  be  constantly  changing.  So  long  as  the  num- 
ber can   be  expressed   in  terms  of  the  unit  one,  it   is 

infinity ;  finite,  and  not  infinite  ;  but  when  the  unit  one  forms 
no  appreciable  part  of  the  number,  the  term  infinite 
is  used  to  express  that  state  of  value,  or  rather,  that 
limit  of  value. 


The  terms, 

how 
employed. 


Are  limits. 


20.  The  terms  zero  and  infinity  are  therefore  em- 
ployed to  designate  the  limits  to  which  decreasing  and 
increasing  quantities  may  be  made  to  approach  nearer 
than  any  assignable  quantity;  but  these  limits  cannot 
be  compared,  in  respect  to  magnitude,  with  any  known 
standard,  so  as  to  give  a  finite  ratio. 

Why  limits?  21.  It  may,  perhaps,  appear  somewhat  paradoxical, 
that  zero  and  infinity  should  be  defined 'as  "the  limits 
of  number  and  space"  when  they  are  in  themselves 
not  measurable.     But  a  limit  is  that  "  which  sets  bounds 

Definition  of  to,  or  circumscribes ;"  and  as  all  finite  space  and  finite 
number  (and  such  only  are  implied  by  the  terms  Space 

of^eprand  and  Number),  are  contained  between  zero  and  infinity, 
we  employ  these  terms  to  designate  the  limits  of  Num- 
ber and  Space. 


Number 


OF     THE     EQUATION. 

22.    The  subject  of  equations   is   divided   into   two 
parts.     The  first,  consists  in  finding  the  equation ;  that 


Subject  of 

equatious  : 

ho\r  divided 

First  part: 


is,  in  the  process  of  expressing  the  relations  existing 


INTRODUCTION. 


IT 


Solution. 


Discussion  ol 
an  equation 


Statement : 
what  it  is. 


between  the  quantities  considered,  by  means  of  the 
algebraic  symbols  and  formula.  This  is  called  the 
Statement  of  the   proposition.     The   second   is   purely     statement. 

Second  part. 

deductive,  and  consists,  in  Algebra,  in  what  is  called 
the  solution  of  the  equation,  or  finding  the  value  of 
the  unknown  quantity ;  and  in  the  other  branches  of 
analysis,  it  consists  in  the  discussion  of  the  equation ; 
that  is,  in  the  drawing  out  from  the  equation  every 
proposition  which  it  is  capable  of  expressing. 

23.  Making  the  statement,  or  finding  the  equation, 
'  is  merely  analyzing  the  problem,  and  expressing  its 
elements  and  their  relations  in  the  language  of  analy- 
sis. It  is,  in  truth,  collating  the  facts,  noting  their 
bearing  and  connection,  and  inferring  some  general 
law  or  principle  which  leads  to  the  formation  of  an 
equation. 

The  condition  of  equality  between  two  quantities 
is  expressed  by  the  sign  of  equality,  which  is  placed 
between  them.  The  quantity  on  the  left  of  the  sign 
of  equality  is  called  the  first  member,  and  that  on 
the  right,  the  second  member  of  the  equation.  Hence, 
an  equation  is  merely  a  proposition  expressed  alge- 
braically, in  which  equality  is  predicated  of  one  quan- 
tity as  compared  with  another.  It  is  the  great  formula 
of  Algebra. 


Equality  of 
two  quanti- 
ties : 
How  ex 
pressed. 
1st  member. 

2d  member. 

Proposition. 


24.  Every  quantity  is  either  abstract  or  concrete : 
hence,  an  equation,  which  is  a  general  formula  for 
expressing  equality,  must  be,  either  abstract  or  con- 
crete. 

2 


Abstract*. 
Concrete^ 


18 


INTRODUCTION. 


Abstract 

equation. 


Concrete 
equation. 


Aii  abstract  equation  expresses  merely  the  relation 
of  equality  between  two  abstract  quantities :  thus, 

a  -jr  b  =  x, 

is  an  abstract  equation,  if  no  unit  of  value  be  assigned 
to  either  member ;  for,  until  that  be  done  the  abstract 
unit  one  is  understood,  and  the  formula  merely  ex- 
presses that  the  sum  of  a  and  b  is  equal  to  x,  and  is 
true,  equally,  of  all  quantities. 

But  if  we  assign  a  concrete  unit  of  value,  that  is, 
say  that  a  and  b  shall  each  denote  so  many  pounds, 
weight,  or  so  many  feet  or  yards  of  length,  x  will  be 
of  the  same  denomination,  and  the  equation  will  be- 
come concrete  or  denominate. 


'Five  opera- 
tions may  be 
performed. 


25.  We  have  seen  that  there  are  five  operations 
which  may  be  performed  on  an  algebraic  quantity 
(Art.  3).  We  assume,  as  an  axiom,  that  if  the  same 
operation,  under  either  of  these  processes,  be  performed 
on  both  members  of  an  equation,  the  equality  of  the 
members  will  not  be  changed.  Hence,  we  have  the 
five  following 


Axioms. 


First. 


Second. 


Third. 


AXIOMS. 

1.  If  equal  quantities  be  added  to  both  members 
of  an  equation,  the  equality  of  the  members  will  not 
be  destroyed. 

2.  If  equal  quantities  be  subtracted  from  both  mem- 
bers of  an  equation,  the  equality  will  not  be  destroyed. 

3.  If  both  members  of  an  equation  be  multiplied  by 
the  same  number,  the  equality  will  not  be  destroyed. 


lNTrlOtrtJCTlOtf. 

4.  If  both  members  of  an  equation  be  divided  by 
the  same  number,  the  equality  will  not  be  destroyed. 

5.  If  the  same  root  of  both  members  of  an  equa- 
tion be  extracted,  the  equality  of  the  members  will 
not   be   destroyed. 

Every  operation  performed  on  an  equation  will  fall 
under  one  or  other  of  these  axioms,  and  they  afford 
the  means  of  solving  all  equations  which  admit  of 
solution, 


l*. 


Fourth. 


Fifth. 


Use  of 
axioms. 


26.    The    term    Equality,   in    Geometry,    expresses     Equality: 

Its  meaning 

that  relation  between  two  magnitudes  which  will  in  Geometry 
cause  them  to  coincide,  throughout  their  whole  ex- 
tent, when  applied  to  each  other.  The  same  term,  its  meaning 
in  Algebra,  merely  implies  that  the  quantity,  of  which 
equality  is  predicated,  and  that  to  which  it  is  affirmed 
to  be  equal,  contain  the  same  unit  of  measure  an 
equal  number  of  times :  hence,  the  algebraic  signifi- 
cation of  the  term  equality  corresponds  to  the  signifi-  Corresponds  w 

.  .  equivalency. 

cation  ot   the  geometrical  term  equivalency. 


27.  We  have  thus  pointed  out  some  of  the  marked 
characteristics  of  Algebra.     In  Algebra,  the  quantities,      classes  of 
about  which  the  science  is  conversant,  are  divided,  as    <iu*ntlt*es  lA 

Algebra. 

has  been  already  remarked,  into  known  and  unknown, 
and  the  connections  between  them,  expressed  by  the 
equation,  afford  the  means  of  tracing  out  further  rela- 
tions, and  of  finding  the  values  of  the  unknown  quan- 
tities in  terms  of  the  known. 


20 


IHTRODTTCTIOW. 


SUGGESTIONS    FOR    THOSE    WHO    TEACH    ALGEBRA. 


Letters  are  but 
mere  symbols. 


Signs  indicate 
operations. 


1.  Be  careful  to  explain  that  the  letters  employed, 
are  the  mere  symbols  of  quantity.  That  of  and  in  them- 
selves, they  have  no  meaning  or  signification  whatever, 
but  are  used  merely  as  the  signs  or  representatives 
of  such  quantities  as  they  may  be  employed  to  denote, 

2.  Be  careful  to  explain  that  the  signs  which  are 
used  are  employed  merely  for  the  purpose  of  indicating 
the  five  operations  which  may  be  performed  on  quan- 
tity ;  and  that  they  indicate  operations  merely,  without 
at  all  affecting  the  nature  of  the  quantities  before  which 
they  are  placed. 

3.  Explain  that  the  letters  and  signs  are  the  ele- 
ments oi'  the  algebraic  language,  and  that  the  language 
itself  arises  from  the  combination  of  these  elements. 

4.  Explain  that  the  finding  of  an  algebraic  formula 
is  but  the  translation  of  certain  ideas,  first  expressed 
in  our  common  language,  into  the  language  of  Algebra  $ 

Its  interpreta-    an(j   that  the  interpretation  of  an  algebraic  formula  is 

tion. 

merely  translating  its  various  significations  into  common 
language. 

5.  Let  the  Language  of  Algebra  be  carefully  studied, 
so  that  its  construction  and  significations  may  be  clearly 
apprehended. 

6.  Let  the  difference  between  a  co-efficient  and  an 
exponent  be  carefully  noted,  and  the  office  of  each  often 
explained  ;  and  illustrate  frequently  the  signification  of 
the  language  by  attributing  numerical  values  to  letters 
in  various   algebraic  expressions. 

7.  Point  out  often  the  characteristics  of  similar  and 


Letters  and 
signs  elements 
of  language. 


Algebraic 
formula 


Language. 


Co-efficient. 
Exponent 


INTRODUCTION. 


21 


dissimilar  quantities,  and  explain  which  may  be  incor- 
porated and  which  cannot. 

8.  Explain  the  power  of  the  minus  sign,  as  shown 
in  the  four  ground  rules,  but  very  particularly  as  it  is 
illustrated  in  subtraction  and  multiplication. 

9.  Point  out  and  illustrate  the  correspondence  be- 
tween the  four  ground  rules  of  Arithmetic  and  Alge- 
bra ;  and  impress  the  fact,  that  their  differences,  where- 
ever  they  appear,  arise  merely  from  differences  in 
notation  and  language :  the  principles  which  govern 
the  operations  being  the  same  in  both. 

10.  Explain  with  great  minuteness  and  particularity, 
all  the  characteristic  properties  of  the  equation ;  the 
manner  of  forming  it ;  the  different  kinds  of  quantity 
which  enter  into  its  composition  ;  its  examination  or 
discussion ;  and  the  different  methods  of  elimination. 

11.  In  the  equation  of  the  second  degree,  be  careful 
to  dwell  on  the  four  forms  which  embrace  all  the  cases, 
and  illustrate  by  many  examples  that  every  equation 
of  the  second  degree  may  be  reduced  to  one  or  other 
of  them.  Explain  very  particularly  the  meaning  of 
the  term  root ;  and  then  show,  why  every  equation  of 
the  first  degree  has  one,  and  every  equation  of  the 
second  degree  two.  Dwell  on  the  properties  of  these 
roots  in  the  equation  of  the  second  degree.  Show  why 
their  sum,  in  all  the  forms,  is  equal  to  the  co-efficient 
of  the  second  term,  taken  with  a  contrary  sign  ;  and 
why  their  product  is  equal  to  the  absolute  term  with  a 
contrary  sign.  Explain  when  and  why  the  roots  are 
imaginary. 


Similar 
quantities. 

Minus  Bign. 


Arithmetic 
and  Algebra 
compared. 


.  Equation. 
Its  properties. 


Equation  of 

the  second 

degree. 


Its  form. 
Its  roots. 


Their  sum. 


Their  product 


Ay?^  of  ma 

IVBE; 


--'V- 


22 


INTRODUCTION. 


General 
Principles 


Should  be 
explained. 


They  lead  to 
feneral  laws. 


12.  In  fine,  remember  that  every  operation  and  rule 
is  based  on  a  principle  of  science,  and  that  an  intelli- 
gible reason  may  be  given  for  it.  Find  that  reason, 
and  impress  it  on  the  mind  of  your  pupil  in  plain  and 
simple  language,  and  by  familiar  and  appropriate  illus-  * 
trations.  You  will  thus  impress  right  habits  of  inves- 
tigation and  study,  and  he  will  grow  in  knowledge. 
The  broad  field  of  analytical  investigation  will  be 
opened  to  his  intellectual  vision,  and  he  will  have 
made  the  first  steps  in  that  sublime  science  which  dis- 
covers the  laws  of  nature  in  their  most  secret  hiding- 
places,  and  follows  them,  as  they  reach  out,  in  omnipo- 
tent power,  to  control  the  motions  of  matter  through 
the  entire  regions   of  occupied   space. 

(See    Davies'    Logic   and    Utility   of    Mathematics, 
Article  Algebra). 


KE  Y  . 


EQUATIONS  OF  THE  FIRST  DEGREE. 

1.  Given  Sx  —  2  -f  24  =  31,  to  find  x. 
Transposing  and  reducing, 

Sx  =  9; 
dividing  both  members  by  3, 

x  =  3. 

2.  Given  x  -f  18  =  Sx  —  5,  to  find  x. 
Transposing  and  reducing, 

-  2x  =  -  23 ; 
dividing  both  members  J>y  —  2, 

x  =  11£. 

3.  Given        6  -  2x  +  10  =  20  -  Sx  —  2,  to  find  *. 
Transposing  and  reducing, 

x  =  2. 

4.  Given  x  -f-  -  x  -f  -  x  =  11,  to  find  a:. 

Multiplying  both  members  by  6.  and  reducing, 

11*  =  66; 
whence,  x  =  6. 


24  KEY  TO  DAVIES'  BOURDON.  [79 

5.  Given  2x  —  -  x  +  1  =  5s  —  2,  to  find  s.    ' 

,*>  , 

Multiplying  both  members  by  2,  transposing  and  reducing, 
-7s  =  -6; 

whence,  x  =  - . 

6.  Given  3as  -f  -  —  3  =  bx  —  a,  to  find  x* 

m 

Multiplying  by  2,  transposing  and  reducing, 

6ax  —  2bx  =  6  —  3a ; 
factoring  the  first  member  of  the  equation,  we  have 

(6a  —  26)  x  sie'e  —  3a ; 
6  —  3a 


whence,  x  = 


6a -26 


7.  Given  i^i  +  |  =  20  -  ?-_!!!. ,  to  find  x. 

Multiplying  both  members  by  6, 

3s  -  9  +  2s  =  120  -  3  «-+  57 ; 
transposing  and  reducing, 

8s  =  186;  .'.  x  =  23$, 

~.  s  +  3   ,   s       i      x  —  5  ^    ,,    , 

8.  Given        —^—  +  3  =  4 J— 1  to  find  * 

Multiplying  both  members  by  12, 

Qx  +  18  +  4s  =  48  -  3s  -f-  15 ; 
transposing  and  reducing, 

13s  =45;'  ,'.  s  =  3^ 


79.]  EQUATIONS  OF  THE  FIRST  DEGREE.  25 

A    ^  ax  —  6   ,    a       6#       bx  —  a 

9.  Given      — h  -  =  — — ,  to  find  x. 

Multiplying  both  members  by  12, 

Sax  —  36  +  4a  =  Gbx  —  \bx  +  4a ; 
transposing,  reducing  and  factoring, 

(3a  -  2b)  x  =  36,  .  • .  *  = 


3a  — 26 


10.  Given  ■ = 4  =/.  to  find  x. 

c  a 

Multiplying  both  members  by  cd, 

3adx  —  2bcx  —  4cd  =fcd ; 
transposing,  reducing  and  factoring, 

n     n«  8MB  — 4       36  —  C 

11.  Given —  =  4  —  6,  to  find  x. 

7  2 

Multiplying  both  members  by  14, 

IQax  -  26  -  216  +  7c  =  56  -  146 ; 

transposing  and  reducing, 

i*  no   i   m      ~  56 +  96 -7c 

16a*  =  56  +  96  —  7c ;  .  • .  *  = 


16a 


12.  Given  -  -  _—  +  -  =  — ,  to  find  z. 

Multiplying  both  members  by  30, 

6x-  10* +  20+  15  x  =  130; 
iransposing  and  reducing, 

liar  =  110;  .-.  x  =10. 


26  KEY  TO  DA  VIES'  BOUEDON.  [80 

X         X         X  X 

13.  Given  TA ;  =/.  to  find  *. 

abc       a 

Multiplying  both  members  by  abed,  and  factoring, 
(bed  —  acd  +  abd  —  abc)  x  =abcdf  .  * .  *  — 


bed  — acd  -f-  abd  —  abc 


14.  Given     x — 1 — - —  =  x  -f-  1,  to  "find  x. 

lo  1 1 

Multiplying  both  members  by  143^ 

143*  -  33a;  -f  55  +  52a;  -  26  —  143*  +  143 ; 
transposing  and  reducing, 

19a?  =  114;  -.  x  =  6. 

15.  Given    5-  §S  -  -^-?  =  -  12ff ,  to  find  x. 

Multiplying  both  members  by  315, 

45*  -  280*  -  63*  4-  189  =  —  3983  ; 
transposing  and  reducing, 

-  298*  ==  -  4172  ;  .  •.  *  =  14. 

16.  Given       2* —  =  — - — ,  to  find  *. 

D  & 

Multiplying  both  members  by  10, 

20*  —8*  +  4  =  15*  — 5; 
transposing  and  reducing, 

—  3*  =  -  9  ;  .  • .  *  =  3. 


80^   86.]  EQUATIONS  OF  THE  FIRST  DEGREE.  27 

bx  ~~  d 

17.  Given       3a;  -\ — -  =  x  +-  a,  to  find  x. 

o 

Multiplying  both  members  by  3, 

9x  +  hx  —  d  ■=  3z  4  3a  ; 
transposing,  reducing  and  factoring, 

3a  +  d 


(6  +  b)  x  =  3a  +  d ; 


6  +  6 


.«    ^.         ("  +  A)  («  —  h)      n         4ab  —  b2        _  a2  —  bx 

18.  Given  •L-L- ^-t -  3a  = — 7 2*4-  — t ; 

a  —  b  a  -}•  6  6 

to  find  #. 

Multiplying  both  members  by  a2b  —  b3,  and  performing  indicated 

operations, 

a2bx  +  2a62.r  -f  &3.r  —  a2b2  —  2ab3  —  b*  -  Sa3b  +  3a&3  =  4a2b2 

—  hab3  4-  b*  —  2a2bx  +  2b3x  +  a4  —  a2bx  —  a262  4-  b3x ; 

transposing,  reducing  and  factoring, 

2b  {2a2  +  ab-  ft*)  a?  =  a4  4-  3a36  4-  4a262  -  Gab3  +  26* ; 

dividing  by  the  co-efficient  of  x, 

_  a4  4-  3a36  4-  4a262  —  Gab3  4-  2b* 
X  ~  2b  (2a2  4-  ab  -  b2) 

STATEMENT  AND  SOLUTION  OF  PROBLEMS. 

8.  Divide  $1000  between  A,  B,  and  C,  so  that  A  shall  have  $72 
more  than  B,  and  C  $100  more  than  A. 

Let         x         denote  the  number  of  dollars  in  B's  share. 
Then  will  x  4-  72     "  "  "  "     A's     " 

and  x  4-  72  4-  100  "  "  "  "     C's     " 

From  the  conditions  of  the  problem, 


28  KEY  TO  DAVIES'  BOURDON.  [86. 

x  +  tf  +  72  4-  x+  172  =  1000;     or,     3*  =  756,     .'.    x  =  252, 
or,         A's  share  is  8324,  B^fc  share  $252  and  C's  share  $424. 

9.  A  and  B  play  together  at  cards.  A  sits  down  with  $84  and  B 
with  $48.  Each  loses  and  wins  in  turn,  when  it  appears  that  A  has 
five  times  as  much  as  B.     How  much  did  A  win? 

Let  x  denote  the  number  of  dollars  that  A  wins. 

Then  will  84  +  x  denote  what  A  has  at  last, 

and  48  —  x  what  B  has  at  last ; 

from  the  conditions  of  the  problem, 

84  +  x  =  5  (48  —  x) ;         or,         84  +  x  =  240  —  5x ; 

whence,  #  —  26 ;         or,         A  wins 


10.  A  person  dying,  leaves  half  of  his  property  to  his  wife,  one 
sixth  to  each  of  two  daughters,  one  twelfth  to  a  servant,  and  the 
remaining  $600  to  the  poor  :   what  was  the  amount  of  his  property  1 

Let         x  denote  the  whole  number  of  dollars  in  the  property. 
Then  will  -       "  "  "  "         "  in  the  wife's  share. 


2 

x 
6 


"  each  daughter's  " 


x 


arid  Z-       "         s.  «  "  "         "  the  servant'*       " 

12 

fromHthe  conditions  of  the  problem, 

multiplying  both  members  by  12,  transposing  and  reducing, 
-  *  =  -  7200 ;         or,         »sc  7200. 


86.]  EQUATIONS  OF  THE  FIRST  DEGREE.  29 

11.  A  father  leaves  his  property,  amounting  to  $2520,  to  four 
sons,  A,  B,  C  and  D.  C  is  to  have  $360,  B  as  much  as  C  and  D 
together,  and  A  twice  as  much  as  B  less  $1000:  how  much  do  A, 
B  and  D  receive  1 

Let        x         denote  the  number  of  dollars  that  D  receives  • 
Then  will  x  +  360       "  "  "  "   B 

and  2x  +  720  -  1000        "  "  "    A        " 

from  the  conditions  of  the  problem, 

360  +  x  +  x+  360  +  2x  +  720  -  1000  =  2520 ; 
transposing  and  reducing, 

Ax  an  2080  ;  .  • .  x  =  520 

or,     D's  share  is  $520  ;  B's  share  $880,  and  A's  share  $760. 

12.  An  estate  of  $7500  is  to  be  divided  between  a  widow,  two 
sons,  and  three  daughters,  so  that  each  son  shall  receive  twice  as 
much  as  each  daughter,  and  the  widow  herself  $500  more  than  all 
the  children :  what  was  her  share,  and  what  the  share  of  each  child  ? 

Let  x  denote  the  number  of  dollars  in  each  daughter's  share  ; 

Then  will  2x         "  *  "  "         son's  " 

and  4x  +  3z  4  500"  "  "         the  widow's  " 

from  the  conditions  of  the  problem, 

4*  -f  3*  4  4s  +  $*  +  500  =  7500 ; 
transposing  and  reducing, 

145?  =  7000;  •.  ar  =  500. 

Daughters'  share  $500;  son's  share  $1000;  widow's  share  $4000. 

13.  A  company  of  180   persons  consists  of  men,   women  ami 


80  KEY  TO  DA  VIES'  BOURDON.  [87. 

children.  The  men  are  8  more  in  number  than  the  women,  and  the 
children  20  more  than  the  men  and  women  together :  how  many  of 
each  sort  in  the  company  ? 

Let  x  denote  the  number  of  women ; 
Then  will  «-f8    "  "  men ; 

and-  x  -f  x  +  8  -f  20  "  children. 

From  the  conditions  of  the  problem, 

#-fa;  +  8-f-a;-fa;  +  8  +  20  =  180; 
transposing  and  reducing, 

4<c  =144;  .'.  x  =  36. 

36  women,  44  men  and  ICO  children. 

14.  A  father  divides  $2000  among  five  sons,  so  that  each  elder 
should  receive  $40  more  than  his  next  younger  brother :  what  is 
the  share  of  the  youngest  1 

Let  x  denote  the  number  of  dollars  in  the  youngest's  share. 


Then  will  x  -f  40  " 

a 

u 

u 

second's 

x  +  80  " 

K 

ft 

1 

third's 

x+  120 

U 

M 

a 

fourth's 

x  -f  160 

(( 

(1 

a 

fifth's 

From  the  conditions  of  the 

probl 

em, 

bx  -f-  400 

-=2000; 

transposing  and  red 

ucing, 

bx  — 

1600; 

. 

' . 

X 

=  320. 

15.  A  purse  of  $2850  is  to  be  divided  among  three  persons,  A, 
B  and  C ;  A's  share  is  to  be  T6T  of  B's  share,  and  C  is  to  have  $300 
more  than  A  and  B  together  :  what  is  each  one's  share  % 


87.]  EQUATIONS  OF  THE  FIRST  DEGREE.  31 

Let         x  denote  the  number  of  dollars  in  B's  share. 

(Sx 
Then  will  —     "       -    "  "  "      A's     U 

and  x  4-  ^  +  300  "  u      C's    " 

From  the  conditions  of  the  problem, 

(\x.  Gx 

x  +  - '  +  x  +  jY  +  300  =  2850; 

clearing  of  fractions,  transposing  and  reducing, 

34#  =  28050  ;  .  • .  x  =  825  ; 

hence,  B's  $825  ;  A's  $450  ;  and  C's  $1575. 

16.  Two  pedestrians  start  from  the  same  point;  the  first  steps 
twice  as  far  as  the  second,  but  the  second  makes  five  steps  while  the 
first  makes  but  one.  At  the  end  of  a  certain  time  they  are  300  feet 
apart.  Now,  allowing  each  of  the  longer  paces  to  be  3  feet,  how  far 
will  each  have  travelled  1 

Let  x       denote  the  number  of  feet  travelled  by  the  first. 

Then  will    |  "  "  steps 

o 


U  ((  u 


■     ££  «  «  m 

3 

j  11        §x        1 hx  ' 

and  1 1  X  — ,  or  -—  "  feet 

■        3  6 

From  the  conditions  of  the  problem, 

~  -  x  =  300  ; 
clearing  of  fractions,  transposing  and  reducing, 


second. 


15* 


9*  =s  1800 ;  .  • .  cc  =  200     and     —  as  500. 

7  6 


32  KEY  TO  DAVIES'  BOtJKDON.  [87 

17.  Two  carpenters,  24  journeymen,  and  8  apprenticec,  received 
at  the  end  of  a  certain  time  $144.  The  carpenters  received  $1  per 
day.  each  journeyman  half  a  dollar,  and  each  apprentice  25  cents : 
how  many  days  were  they  employed  ? 

Let        x  denote  the  number  of  days. 

Then  will  x       "  "  dollars  due  each  carpenter. 

x 


«               11                         ti 
2 

u 

journeyman. 

and           %      "            " 
4 

f 

apprentices ; 

from  the  conditions  of  the  problem, 

• 

reducing, 

144; 

16*  ==144; 

X 

=  9. 

18.  A  capitalist  receives  a  yearly  income  of  $2940  ;  four  fiths  of 
his  money  bears  an  interest  of  4  per  cent.,  and  the  remainder  of  5 
per  cent. :  how  much  has  he  at  interest  % 

Let      x     denote  the  number  of  dollars  at  interest. 

4x         4 

Then  will  —  X  r^r  denote  the  interest  of  1st  parcel. 

x         5 
ana  5  X  100 

From  the  conditions  of  the  problem, 

4%        4        x        5         ™.,v 

TxToo  +  5xIoo=:2940; 

clearing  of  fractions,  and  reducing, 

21z  sb  1470000 ;  .  • .  x  =  70000. 


87.]  EQUATIONS  OF  THE  FIRST  DEGREE.  33 

10.  A  cistern  containing  60  gallons  of  water  has  three  unequal 
cocks  for  discharging  it ;  the  largest  will  empty  it  in  one  hour,  the 
second  in  two  hours,  and  the  third  in  three  :  in  what  time  will  the 
cistern  be  emptied  if  they  all  run  together  ? 

Let  x  denote  the  required  number  of  minutes. 
Then  since  the  first  emits   1   gallon  per  minute,  the  second  \  of  a 
gallon  per  minute,  and  the  third  J  of  a  gallon, 

x  will  denote  the  number  of  gallons  emitted  by  the  1st. 
x 


3d. 


From  the  conditions  of  the  problem, 


clearing  of  fractions  and  reducing, 

11*  =  360  .-.  x  =  32j*jm. 

20.  In  a  certain  orchard  £  are  apple-trees,  -J-  peach-trees,  4  plum- 
trees,  120  cherry-trees,  and  80  pear-trees  :  how  many  trees  in  the- 
orchard  1 

Let  x  denote  the  whole  number  of  trees. 


Then  will    ^      " 

u 

a 

"  apple-trees. 

X 

4 

H 

<( 

"  peach-trees. 

6 

ii 

u 

"  plum-trees. 

From  the  conditions  of  the 
3 

problem, 

34  KEY  TO  DAVIES'  BOURDON.  L87-88. 

1+5  +  1+120  +  80  =  *; 

clearing  of  fractions,  transposing  and  reducing, 

—  *  =  -  2400  .  • .  x  =  2400. 

21.  A  farmer  being  asked  how  many  sheep  he  had,  answered  that 
he  had  them  in  five  fields;  in  the  1st  he  had  -J,  in  the  2d  -J,  in  the 
3d  J,  in  the  4th  J^  and  in  the  5th  450  :  how  many  had  he  % 

Let  x  denote  the  whole  number  of  sheep : 

Then  will  %         "  "  "  "      in  1st  field. 


2d     " 
3d     " 


and  ^        «  "  "  "  4th    " 

1,2 

From  the  conditions  of  the  problem, 

x      x      x       x 

multiplying  both  members  by  24,  transposing  and  reducing, 
-  9s  =  —  10800  .  * .  x  =  1200. 

22.  My  horse  and  saddle  together  are  worth  $132,  and  the  horse 
is  worth  ten  times  as  much  as  the  saddle :  what  is  the  value  of  the 
horse I 

Let        x  denote  the  number  of  dollars  that  the  saddle  is  worth. 

Then  will  10s    "  "  "  "      horse 

From  the  conditions  of  the  problem, 


88.]  EQUATIONS  OF  THE  FIRST  DEGREE.  35 

*  +  10*  =  132; 
reducing,     11*  =  132         .  •.      *  =  12;     whence,     10*=  120. 

23.  The  rent  of  an  estate  is  this  year  8  per  cent,  greater  than  it 
was  last.     This  year  it  is  $1890 :  what  was  it  last  year  1 

Let        *  denote  the  number  of  dollars  in  last  year's  rent. 
Then  will* -h^  "  "  "  this     "        tt 


From  the  conditions  of  the  problem, 

_8* 
100 


*  +  T7^=  1890; 


clearing  of  fractions  and  reducing, 

108*  x=  189000;  .-.  *  =  1750. 

24.  What  number  is  that  from  which,  if  5  be  subtracted,  J  of  the 
remainder  will  be  40  ? 

Let  *  denote  the  number  required : 

From  the  conditions  of  the  problem, 

#(*-5)  =  40; 

clearing  of  fractions,  performing  operations  indicated,  transposing 
and  reducing, 

2x  =  130 ;  .  • .  x  =r  65. 

25.  A  post  is  J  in  the  mud,  £  in  the  water,  and  ten  feet  above  the 
water :  what  is  the  whole  length  of  the  post  ? 

Let  x  denote  the  number  of  feet  in  length. 

* 


Then  will    -         «  «  «  «  the  mud;  < 

\ 


4  "" " » 


and  |         «  «  "  «         the  water: 


36  KEY  TO  DA  VIES'  BOITRDOtf.  [88. 

From  the  conditions  of  the  problem ; 

x      x 
2+3+10  =  *; 

clearing  of  fractions,  transposing  and  reducing, 

-5z=-120;  .-.  a:  =  24. 

26.  After  paying  l  and  A  of  my  money,  I  had  66  guineas  left  in 
my  purse  :  how  many  guineas  were  in  it  at  first  ? 

Let  x  denote  the  number  at  first ; 

from  the  conditions  of  the  problem, 

x      x 
4       5 

clearing  of  fractions,  transposing  and  reducing, 

liar  =  1320;  .-.  x  =  120. 

27.  A  person  was  desirous  of  giving  3  pence  apiece  to  some 
beggars,  but  found  he  had  not  money  enough  in  his  pocket  by  8 
pence;  he  therefore  gave  them  each  two  pence  and  had  3  pence 
remaining  :  required  the  number  of  beggars. 

Let  x  denote  the  number  of  beggars; 
then,  by  the  first  condition, 

Zx  —  8         denotes  the  number  of  pence  j 
by  the  second  condition, 

2x  -f  3         denotes  the  number  of  pence  j 
hence,  3z  —  8  =  2x  -f  3  ; 

transposing  and  reducing, 

*=11. 


88.]  EQUATIONS  OF  THE  FIR6T  DEGREE.  37 

28.  A  person  in  play  lost  -J-  of  his  money,  and  then  won  3 
shillings  ;  after  which  he  lost  j-  of  what  he  then  had  ;  and  this  done, 
found  that  he  had  but  12  shillings  remaining:  what  had  he  at  first? 

Let  x        denote  the  number  of  shillings  at  first ; 

Then  will  x  —  -  +  3       "  "  "       after  first  sitting ; 

and  ^_!+3)-i(s-!  +  3) 

will  denote  what  he  finally  had  ; 

hence,  from  the  conditions  of  the  problem, 

*-|  +  3-l(*-|+3)  =  12; 

clearing  of  fractions,  performing  indicated  operations,  transposing 
and  reducing, 

6*  =  120;  .-.  s  =  20. 

29.  Two  persons,  A  and  B,  lay  out  equal  sums  of  money  in  trade ; 
A  gains  $126,  and  B  loses  $87,  and  A's  money  is  now  double  B's : 
what  did  each  lay  out  1 

Let  x  denote  the  number  of  dollars  laid  out  by  each ; 

Then  will    x  +  126       "  "  "      A  had ; 

and  s  —  87         "  "  "      B     " 

From  the  conditions  of  the  problem, 

x  -f  126  =  2  (x  -  87)  ; 
performing  indicated  operations,  transposing  and  reducing, 
—  x  -  —  300  ;  .  • .  x  =  300. 

30.  A  person  goes  to  a  tavern  with  a  certain  sum  of  money  in  his 


38  KEY  TO  DAVIES'  BOURDON.  [88-89. 

pocket,  where  he  spends  2  shillings;  he  then  borrows  as  much 
money  as  he  had  left,  and  going  to  another  tavern,  he  there 
spends  2  shillings  also ;  then  borrowing  again  as  much  money  as 
was  left,  he  went  to  a  third  tavern,  where,  likewise,  he  spent  2 
shillings  and  borrowed  as  much  as  he  had  left ;  and  again  spend- 
ing 2  shillings  at  a  fourth  tavern,  he  then  had  nothing  remaining. 
What  had  he  at  first  \ 

Let  x  denote  the  number  of  shillings  at  first. 
Then,  from  the  first  condition, 

x  — -    2                                will  denote  what  he  has  after  1st  visit. 

2(x-    2)-2or2z-    6         "            "  "         "   2d     « 

2(2*  —    6)—  2  or  4x  —  14         "             «•  «        "   3d     " 

2  (4r  - 14)  —  2  or  Sx  —  30         "             "  "         "    4th    " 
From  the  conditions  of  the  problem, 

8z  -  30  =  0     or    8*  =  30 ;             .  • .  x  =  3  J . 
or  the  amount  at  first  was  3s.  9d. 

31.  A  farmer  bought  a  basket  of  eggs,  and  offered  them  at  7  cents 
a  dozen.  But  before  he  had  sold  any,  5  dozen  were  broken  by  a 
careless  boy,  for  which  he  was  paid.  He  then  sold  the  remainder  at 
8  cents  a  dozen,  and  received  as  much  as  he  would  have  got  for  the 
whole  at  the  first  price.     Eow  many  eggs  had  he  in  his  basket  ? 

Let        x  denote  the  number  of  dozens  at  first ; 

Then  wills  -  5%     "  "  "       sold; 

and  8  (x  —  5) "  "  cents  received ; 

Ix  "  "  "    first  asked; 

hence,  Ix  =  8  (x  —  5)  ;  .  • .  x  =  40. 


>.]  EQUATIONS  OF  THE  FIBST  DEGREE. 

SIMULTANEOUS  EQUATIONS  OF  THE  FIRST  DEGREE. 

( 2x  +  3y  =  16 ) 

1.  Given  1  j-to  find  x  and  y. 


39 


3*  — 2# 

Multiply  both  members  of  the  first  by  2,  and  of  the  second  by  3  ; 
4x 
9x 

whence,  by  addition,  member  to  member,  we  have, 

13*  =  65  ;  .  * .  x  =  5,         also,        y  =  2. 


+  6#  =  32 ) 
-6y  =  33) 


2.  Given 


2^      3y  _  9 
T+1f  ~  120J 


.    to  find  x  and  y. 


Clearing  of  fractions,  and  then  multiplying  both  members  of  the  first 
by  16,  and  of  the  second  by  5, 

128*  +  240y  =  144) 
450*  -f-  240y  =  305 ) 

whence,  by  subtracting,  member  from  member, 
322*  =  161 ; 


*  =  ~,     also,  by  substitution,  y  =  -• 
Z  3 


3.  Given 


+  7y  =  99 
+  7*  =  51 


►  to  find  x  and  y. 


Multiplying  the  first  by  343  and  the  second  by  7 ; 

49*  +  2401*/  =  33957  ) 
y  +    49*    =    357    j 


^>  0?  TH5     X  ■£ 


^  *>. 


40 


KEY  TO  DAVIES    BOURDON. 


[95. 


by  subtraction, 
2400y  =  33600; 


y  =  14 ;     also,  by  substitution,    x  =  7. 


4.  Given      < 


|-12  =  |+8 


;  to  find  x  and  y. 


Clearing  of  fractions  and  transposing, 

2x  —  y  =  80 
47z  -18y  =  2100; 

multiplying  both  members  of  the  first  by  18,  and  subtracting  the 
result  from  the  second,  member  from  member, 
11  x  sa  660 ;         .  • .         x  =  60  ;         by  substitution,        y  =  40. 


'x+    y+    2  =  29     •     •     •     (1)" 

5.  Given  * 

z  +  2y  +  3z  =  62     •     •     •    (2) 
1+    |+   |=10     •    •    •    (3) 

-  to  find  x,  y  and  *. 

Combining  (1)  and  (2), 

y  +  2z  =  3?     •     •     •     •     (4) ; 

combining  (1)  and  (3) 

2y  +  3*  =  54     ....     (5); 

combining  (4)  and  (5) 

z  =  l 

2 ;     by  successive  substitutions,   x 

=  8,     y  =  9. 

{2x  +  4y-2z  =  22    •     -     (1)] 
6.  Given  -j  Ax  —  2y  +  5z  =  18     •     •     (2)  [\ 
[6x  +  7y  —    z  =  63 

Combining  (1)  and  (2), 


•  •    (2)[; 

•  •    (»)J 


95.] 


EQUATIONS   OF   THE   FIRST   DEGREE. 


41 


lOy-  110  =  26    ...     (4); 
combining  (1)  and  (3), 

5y  —  8z  =    3     .     .     .     (5)  ; 
combining  (4)  and  (5), 

5*  =  20;  .-.         «  =  4. 

By  successive  substitutions,  x  =  3,        y  =  7. 


,  +  |  +  3  =  32 

7.  Given              < 

..  ;  to  find  a;,  y  and  #. 

Clearing  of  fractions, 

6a;  -f-    3y  +    2z  =  192 

•  •  :  (i); 

20a;  +  15y  +  12*  =  900 

•    •    •    (2); 

15a;  +  12y  +  10z  =  720 

.    .    .     (3); 

combining  (1)  and  (2), 

16a?  +  3y*=  252    . 

•   •    (4); 

combining  (1)  and  (3), 

15a;  +  %  =  240 

•   •    (5); 

combining  (4)  and  (5), 

x  =  12; 

by  successive  substitutions,    y  =  20,     z 

=  30. 

8.  Given  - 

7a;- 
4y- 

5y  — 
4y- 

2z  +  3w  =  17     . 
2z  +      *=11     . 
3a;  —  2t*  ==    8     . 

3?/  +  2*  =    9     . 
30  +  8w  =  33     . 

•  .  (1)1 

•  •     (2) 

•  •     (3) 

•  •     (4) 
.     .     (5) 

►  ;  to  find  x,  yt 
2,  t  and  u. 

42  KEY  TO  DA  vies'  bouedon.  [98-99. 

Combining  (2)  and  (4), 

4y-4z  +  3u=  13     .     .     .     (6); 
combining  (1)  and  (3), 

35y  -  6z  -  hu  =  107     .     .     .     (7) ; 
combining  (5)  and  (6), 

12y  +  41w  =  171     .     .     .     (8); 
combining  (5)  and  (7), 

35y +11^  =  173    .    .     .     (9); 
combining  (8)  and  (9), 

1303w  =  3909;  .-.         «  =  3; 

by  successive  substitutions,      x  =  2,     y  =  4,     2  =  3,     <  r=  1. 

PROBLEMS  GIVING  RISE  TO  SIMULTANEOUS  EQUATIONS  OF  THE 
FIRST  DEGREE. 

5.  What  two  numbers  are  they,  whose  sum  is  33  and  whose 
difference  is  71 

Let  x  denote  the  first,  and  y  the  second. 
From  the  conditions, 

x  -f  y  =  33 
x-y  =    7; 
whence,  by  combination, 

x  =  20,  y  —  13. 

6.  Divide  the  number  75  into  two  such  parts,  that  three  times  f,n« 
greater  may  exceed  seven  times  the  less  by  15. 

Let  x  denote  the  greater,  and  y  the  less 
From  the  conditions  of  the  problem, 


EQUATIONS    OF   THE   FIKST  DEGBEE.  *       43 

X  +     y  =  75 

Sx-7y  =  l5; 

by  combination,     lOy  =  210  ;         .-.     y  =  21 ;     also,     a?  =  54. 

7.  In  a  mixture  of  wine  and  cider,  %  of  the  whole  plus  25  gallons 
was  wine,  and  -J  part  minus  5  gallons,  was  cider  :  how  many  gallons 
were  there  of  each  1 

Let        x  denote  the  number  of  gallons  of  wine  ; 


and            y 

"                 "            cider ; 

then  will    x  +  y 

"             ♦  "            mixture. 

From  the  conditions, 

x-p  +  *  =  - 

H^tm 

clearing  of  fractions, 

transposing  and  reducing, 

y  —  x  =  —  50 

—  2y  +  x  =3       15 

by  combination, 

y 

=  35  ;         and         x  =  85 

8.  A  bill  of  £120  was  paid  in  guineas  and  moidores,  and  the 
number  of  pieces  of  both  iorts  that  were  used  was  just  100;  if  the 
guineas  were  estimated  at  21s.,  and  the  moidores  at  27s.,  how  many 
were  there  of  each  1 

Let     x  denote  the  number  of  moidores ; 

and         y         "  "  "  guineas ; 

then,  since  £120  =  2400s.,  we  have,  from  the  conditions, 


44      "  KEY  TO  DAVIES'  BOURDON.  [99 

X  +        ?/=      100 

27*  +  21y  =  2400  ; 
by  combination,     6y  =  300 ;     .  * .     y  =  50 ;     also,     a;  =  50. 

9.  Two  travellers  set  out  at  the  same  time  from  London  and 
York,  whose  distance  apart  is  150  mile*  ;  they  travel  toward  each 
other ;  one  of  them  goes  8  miles  a  day,  and  the  other  7 ;  in  what 
time  will  they  meet] 

Let  x  denote  the  number  of  miles  travelled  by  the  first ; 

y        "  "  "  "  "        second; 


then  will     f         "  "  days         "  "       first ; 

o 

and  |        "  "  «  "  "        second; 

From  the  conditions, 


whence,  by  combination, 


*  +  y  =  150; 
x       y 
8=7; 


x 


x  =  80        and        -  as  10,         the  number  of  days. 

10.  At  a  certain  election,  375  persons  voted  for  two  candidates; 
and  the  candidate  chosen  had  a  majority  of  91 ;  how  many  voted  for 
each? 

Let  x  denote  the  number  of  votes  received  by  the  first; 
y        "  "  "  "  "      second; 

from  the  conditions  of  the  problem, 

x  4-  y  =  375 
*  =  y+    91; 
by  combination,  x  =  233,        y  =  142. 


99.]  EQUATIONS  OF  THE  FIRST  DEGREE.  45 

.    11.  A's  age  is  double  B's,  and  B's  is  triple  C's,  and  the  sum  of 
all  their  ages  is  140  :  what  is  the  age  of  each  % 
Let  x  denote  the  age  of  A 
y         «  "  B 

z        "  "  C 

from  the  conditions  of  the  problem) 

x~    >2y     .     •     •     •     (1); 
y=    Zz     .     .     .     •     (2); 
x  +  y+z  =  U0     .     •     .     i     (3); 
from     (1)  and  (2),  x  =  6z  ; 

substituting,  y  =:  3*,  and     a;  ss  6z,    in  (3),  and  reducing, 

10*  =  140 ;  .  • .  z  =c  14,  .  .#  as  84,     y  =  42. 

12.  A  person  bought  a  chaise,  horse  and  harness,  for  £60 ;  the 
Horse  came  to  twice  the  price  of  the  harness,  and  the  chaise  to  twice 
the  price  of  the  horse  and  harness :  what  did  he  give  for  each  1 
Let  x  denote  the  number  of  pounds  paid  for  the  harness ; 
y         "  "  "  "         "       horse; 

z         "  "  "  "         "       chaise; 

from  the  conditions  of  the  problem, 

y  =  2x     •     ♦     '     •     (1) 

z  =  2(x  +  y)     •     •     •     •     (2) 

*  +  #■$-   *mM    ■•    *    •    •  ;(8) 

from  (2)     and     (1)     z  =  Qx; 

substituting         z  sb  Gx      and      y  =.2x      in     (3) 

9#  ss  60,     .  * .   x  B=  6|L     also,  by  substitution,   y  =  13 J,     2  =  40; 

hence,  the  price  of  the  chaise  was  £40;  of  the  horse  £13  6*.  Sd. ; 
and  that  of  the  harr  .^ss  £6  1 3s.  4c?. 


46  KEY  TO  DAVIE8*  BOURDON.  [99-100. 

13.  A  person  has  two  horses,  and  a  saddle  worth  £50 ;  now,  if- 
the  saddle  be  put  on  the  back  of  the  first  horse,  it  will  make  his 
value  double  that  of  the  second  ;  but  if  it  be  put  on  the  back  of  the 
second,  it  will  make  his  value  triple  that  of  the  first :  what  is  the 
value  of  each  horse  ? 

Let     x     denote  the  number  of  pounds  the  1st  horse  is  worth; 

y         "  "  «  "     2d       "  " 

from  the  conditions  of  the  problem, 

x  +  50  =  2y 

y  -f  50  =  Zx ; 
whence,  by  combination, 

x  -    30     y  =    40. 

14.  Two  persons,  A  and  B,  have  each  the  same  income.  A  saves 
J  of  his  yearly;  but  B,  by  spending  £50  per  annum  more  than  A, 
at  the  end  of  4  years  finds  himself  £100  in  debt ;  what  is  the  income 
of  each? 

Let     x    denote  the  number  of  pounds  in  the  income  of  A  ;     . 
y         u  «  "  "         "  B; 

by  the  conditions  of  the  problem,  these  are  equal ;  one  only  will  h* 

used.     Then  will 
4 

-x     denote  what  A  spends  per  year : 
5 

ic +50"       "    B       " 
5 

from  the  conditions  of  the  problem, 

4^  +  5o)  -.=  4x+  100; 

whence,  performing  indicated  operations,  transposing  and  reducing, 
4a;  =    500         .  * .  x  =     125. 


100.]  EQUATION8  OF  THE  FIRST  DEGREE.  47 

15.  To  divide  the  number  36  into  three  such  parts,  that  J  of  the 
first,  £  of  the  second,  and  J  of  the  third,  may  be  all  equal  to  each 
other. 

Let  #,  y   and  r,  denote  the  parts. 
From  the  conditions  of  the  problem, 

#  +  y  4-  x  =  36 

*  _  t 
2~3 

a;  _  z 
2~4; 

clearing  of  fractions,  and  combining, 

9x  =  72     .  * .     ar  =  8  ;     whence,     y  va  12     and     «  =  16. 

16.  A  footman  agreed  to  serve  his  master  for  £8  a  year  and 
livery,  but  was  turned  away  at  the  end  of  7  months,  and  received 
only  £2  135.  4d.  and  his  livery  :  what  was  its  value? 

Let  x  denote  the  value  of  livery,  expressed  in  shillings :  £8  =2 
160*.,  and  £2  13*.  Ad.  =  53£s. ; 

Then  will  ( — — — I  denote  the  value  of  wages  1  month, 

and  7(i^  +  _*)       u  .  .,       7 

by  the  conditions  of  the  problem, 
/160  +  x\ 

7(-T2-)  =  53*  +  * 

1120  +  7x  =  640  +  12  x 
—  5x  =  —  480 
*  =  96  .'.  value,  £4.16*. 


48  KEY  TO  DAVIE8    BOtTBDON.  [100- 

i 

17.  To  divide  the  number  90  into  four  such  parts,  that  if  the  first 
be  increased  by  2,  the  second  diminished  by  2,  the  third  multiplied 
by  2,  and  the  fourth  divided  by  2,  the  sum,  difference,  product, 
and  quotient,  so  obtained,  will  be  all  equal  to  each  other. 

Let     #,  y,  z    and  «,  denote  the  parts; 
from  the  conditions  of  the  problem, 

x  +  y  +  z-\-u  =  Q0 
x  +  2z=y  —2 
x  +  2  ss  2z 
u 

whence  we  find  from  the  last  three  equations, 

x 
y  =  x  -f  4,     2^-rl,     and     u  =  2x  -f*  4 ; 

substituting  these  values  in  the  first  equation, 

flj  +  «  +  4+|+l+2aj  +  4  =  90;or44«  =  81;.-.    x  =  18 ; 
whence,  by  substitution,     y  —  22,     z  —  10,     and     u  =  40. 

18.  The  hour  and  minute  hands  of  a  clock  are  exactly  together  at 
12  o'clock  :  when  are  they  next  together. 

1st  Solution. 
Let  x  denote  the  number  of  minute  spaces  passed  by  the  hour 
hand  before  they  come  together ;         *     * 
and  y  the  number  passed  by  the  minute  hand  ; 
then,  since  the  latter  travels  12  times  as  fast  as  the  former,  and 
since  it  has  to  gain  60  spaces,  we  have, 

x  —  y  m  60 

*  =  12y; 


100  J  EQUATIONS  OF  THE  FIRST  DEGREE.  49 

by  combination, 

lly  =  60  .  • .  y  =  5T5T ;      also,      x  =  65T5T  ; 

hence,  they  will  be  together,  65-^j-  minutes  after  12  o'clock,  or  at  1 
o'clock,  T5T  minutes,  and  at  the  end  of  every  succeeding  equal 
portion  of  time. 

2d  Solution. 
The  minute  hand  will  pass  the  hour  hand  11  times  before  they 
again  come  together  at  12  o'clock,  and  the  times  between  any  two 
consecutive  coincidences  will  be  equal.     Hence  each  time  will  be 
equal  to  12  hours  divided  by  11  =  1-^jhr.  =  lkr.  b^m. 

19.  A  man  and  his  wife  usually  drank  out  a  cask  of  beer  in  12 
days ;  but  when  the  man  was  from  home,  it  lasted  the  woman  30 
days ;  how  many  days  would  the  man  be  in  drinking  it  alone  ? 

Let     x    denote  the  number  of  days  it  takes  the  man  to  drink  it ; 
y        "  "  "  "  woman  «        " 

then,  if  the  whole  quantity  of  beer  be  denoted  by  1, 

-  will  denote  the  quantity  drank  by  the  man  in  1  day  ;     and 
x 


_                    M                                 li                    M              M 

y 

woman  j 

from  the  conditions  of  the  problem, 

x^  y       12 

1        1 

y~30; 

substituting  the  value  of  -  in  the  first  equation. 

1+1-1. 
x  ^  30  ~  12 ' 

50  KEY  TO  DAVIES'  BOURDON.  [100. 

clearing  of  fractions, 

60  -f  2a?  =  5a?  ;  .  • .  a?  =  20. 

20.  If  A  and  B  together  can  perform  a  piece  of  work  in  8  days, 
A  and*  C  together  in  9  days,  and  B  and  C  in  10  days  :  how  many 
days  would  it  take  each  person  to  perform  the  same  work  alone  1 

Let  the  work  be  denoted  by  1  ; 

Let     a?     denote  the  work  done  by  A  in  one  day  ; 

y  U  «  U  g       «  u 


then  will  -,  -   and  -  respectively  denote  the  number  of  days  that 
it  will  take  A,  B,  and  C  severally  to  do  the  work ; 


from  the  conditions  of  the  problem, 

x  +  y  =    |     .     .     .     . 

0) 

x+z=    J     .     .     .     . 

(2) 

y  +  z  =  iV   •   •   •   • 

(3); 

clearing  of  fractions, 

8a?  -f-    8y  =  1     •     •     »     • 

(4)        1 

9a?  +    9z  =  1     •     •     •     • 

(5) 

lOy  +  102  =  1     •     •     •     • 

(6); 

combining 

(4)     and     (5), 

72y~72z  =  l     •     •     •     • 

(V, 

combining 

(6)     and     (7), 

1440y  =  82 

y  =  Tiki 

substituting 

in     (1)     and     (3), 

a? : 

—  T  ~~  TJTcT  =  Tfo  \           z  ==  T  O 

~    720  =  73 

hence, 

i  =  14Ji;         -=17Jf5 
x           49            y           41 

\-*a 

100— 101.]  EQUATIONS  OE  THE  FIRST  DEGREE.  51 

21.  A  laborer  can  do  a  certain  work  expressed  by  a,  in  a  time 
expressed  by  b  ;  a  second  laborer,  the  work  c  in  a  time  d ;  a  third, 
the  work  e  in  a  time/.  Required  the  time 'it  would  take  the  three 
laborers,  working  together,  to  perform  the  work  g  ? 

If  a  laborer  can  do  a  piece  of  work  denoted  by  a,  in  a  number  of 
days  denoted  by  b,  he  can  do  in  1  day  so  much  of  the  work  as  is 

denoted  by  j  ;  the  second  in  1  day  can  do  so  much  as  is  denoted  by 
-;  and  the  third  so  much  as  is  denoted  by-;    hence,    the   three 

u  J 

working  together  can  do 

a       c       e  _  adf  ■+•  bcf  -f  bde 
6  +  d     f=  bdf 

Let  x  denote  the  time  required  to  perform  the  work  g ;    then, 

the  three  can  perform  the  work  -  in  the  time  1 :  * 

r  x 

from  the  conditions  of  the  problem, 

p  __  adf  4-  bcf  -f  bde 
x~  bdf 

taking  the  reciprocals  of  each  member,  and  then  clearing  of  fractions, 

we  have, 

_  bdfg 

X '^  adf±bcj+b<k 

In  this  example  only  a  single  unknown  quantity  has  been  used, 
and  it  may  be  remarked  that  many  other  examples,  in  this  chapter, 
may  be  more  easily  solved  by  a  single  unknown  quantity  ;  in  such 
ca»es  more  than  one  has  been  used  for  the  purpose  of  illustration. 

22.  If  32  pounds  of  sea  water  contain  1  pound  of  salt,  how  much 
fresh  water  must  be  added  to  these  32  pounds,  in  order  that  the 


fTJHIVBUS: 


5£  KET  TO  DAVIES*  BOTTKDON.  [101. 

quantity  of  salt  contained  in  32  pounds  of  the  new  mixture  shall  be 
reduced  to  2  ounces,  or  A  of  a  pound  % 

Let  x  denote  the  number  of  pounds  to  be  added  ; 

then  will  — - - — -   denote  the  number  of  pounds  of  salt  in  each 
32  +  x  r 

pound  of  the  mixture,  but  this  we  know  to  be  —  x  ~,  or  *- — ►  j 

O/Z       o  /Zoo 

hence  from  the  conditions  of  the  problem, 

32T^  =  i     °r'     33  +  *  =  256.    OT>    *  =  224- 
This  problem  is  also  solved  by  a  single  unknown  quantity  more 
readily  than  by  two. 

23.  A  number  is  expressed  by  three  figures  j  the  sum  of  these 
figures  is  11;  the  figure  in  the  place  of  units  is  double  that  in  the 
|jlace  of  hundreds  ;  and  when  297  is  added  to  this  number,  the  sum 
obtained  is  expressed  by  the  figures  of  this  number  reversed. 
What  is  the  number  ? 

Let  x,  y    and  z  denote  the  digits  in  their  order  j 
then  will  the  number  be  denoted  by 

100* -f  \§y+z\ 
from  the  conditions  of  the  problem, 

x+       y  +  z=zll      <     <     -     • (1) 

z  =    2* «     «     < (2) 

100*  +  lOy  +  z  -f  297  ss  100*  +  lOy  -f  x     -     •     •     (3)} 
reducing  (3),  gives 

99*  ~  99*  =  297 (4); 

substituting  *  =  2*  in  (4),  and  reducing, 

99*  =  297;  .-.  *  =  3 ; 


101."]  EQUATIONS  OF  THE  FIRST  DEGREE.  53 

whence,  by  successive  substitutions, 

y  =  2,         2  =  6.  ^4ns.  326. 

24.  A  person  who  possessed  $100000  dollars,  placed  the  greatei 
part  of  it  out  at  5  per  cent,  interest,  and  the  other  part  at  4  pe) 
cent.  The  interest  which  he  received  for  the  whole  amounted  t* 
4640  dollars.     Required  the  two  parts. 

Let     x    denote  the  greater  part ; 
y         "         "  lesser       " 
From  the  conditions  of  the  problem, 

I* 

— —  =  interest  on  x  dollars  at  5  per  cent. ; 
100  » 

i|=         "  y         «  4     «       «        then, 

x  +  y  =  100000     ....     (1) 

^  +  iy=4640     ....     (2) 
100  ^  100  ™ 

clearing  (2)  of  fractions, 

5x  +  4y  =  464000     ....     (3) 
combining  (1)    and     (3), 

y  =  36000,         whence         x  =  64000. 

25.  A  person  possessed  a  certain  capital,  which  he  placed  out  at 
a  certain  interest.  Another  person  possessed  10000  dollars  more 
than  the  first,  and  putting  out  his  capital  1  per  cent,  more  advan- 
tageously, had  an  income  greater  by  800  dollars.  A  third,  possessed 
15000  dollars  more  than  the  first,  and  putting  out  his  capital  2  per 
cent,  more  advantageously,  had  an  income  greater  by  1500  dollars. 
Required  the  capitals  and  the  three  rates  of  interest. 


54:  KEY  TO  DAVIES*  BOURDON.  [101. 

Let  x  denote  the  number  of  dollars  in  1st  capital; 

and      y  the  rate  per  cent. ;  then, 

(x  x  v^ 

-i -^  will  denote  the  number  of  dollars  of  1st  income ; 

2d        " 

3d        " 


(*+ 10000)  (y-H) 

100                                       v 

(1            u 

(*+15000)(y+2) 
100 

ll           u 

from  the  conditions  of  the  problem, 

(x  +  10000)  (y  +  1) 
100 

x  x  y 
100 

+  800 

(x  +  15000)  (y  +  2)   * 
100 

x  x  y 
100 

+  1500; 

clearing  of  fractions,  performing  indicated  operations,  transposing 
and  reducing,  v 

lOOOOy  -f    *  =    70000 
15000y-f  2*  =  120000; 
combining  and  reducing, 

5000y  =  20000         .  • .        y  =  4 ;  and 

Oy  substitution,  a;  =  30000 ; 

2d.  $40000,     rate  5  per  cent. 
3d.  $45000,     rate  6     "       " 

26.  A  cistern  may  be  filled  by  three  pipes,  A,  B,  C.  By  the  two 
first  it  can  be  filled  in  70  minutes ;  by  the  first  and  third  it  can  be 
filled  in  84  minutes ;  and  by  the  second  and  third  in  140  minutes. 
What  time  will  each  pipe  take  to  do  it  in  ?  What  time  will  be 
required,  if  the  three  pipes  run  together  ? 

Call  the  contents  of  the  cistern  1. 


101.]  EQUATIONS  OF  THE  FIKST  DEGREE.  55 

Let  x  denote  the  quantity  discharged  in  1  minute  by  the  first  ; 
y        "  "  "  "         "         "       second; 

z         "  "  "  "         «         «       third; 

then  will   -.     -     and    -     denote  the  number  of  minutes  required 
x      y  z 

for  the  pipes,  separately,  to  fill  the  cistern  ;         and, 

1 


x  4-  y  +  2' 

will  denote  the  number  of  minutes  required  for  all  three  tc  fill  it, 

running  together ; 

from  the  conditions  of  the  problem, 

x+y=ro  •  ■  • 

•     (1) 

*+v=|  •  •  • 

•     (2) 

y  +  2  =  no-  '•■  ' 

•   (3); 

clearing  of  fractions, 

70*  +    70y  =  1    •     •    • 

•     (4) 

84z  +    84z  =  1    •    •     • 

•     (5) 

140y  +  140s  =  1    -     •    • 

•    (6); 

combining  (1)  and  (2), 

840*/  -  840^  r=  2    •     .     t     •     (7)  ; 
combining  (6)  and  (7), 

1680y  =  8,  .'.  V  =  tL 


210' 


substituting  in  (1),  and  transposing, 


_L    _L    JL    _L 

*  "  70      210  ~"  210  ~  105  ; 


56  KEY  TO  DAVIES'  BOURDON.  [102. 

substituting  in    (3), 

_  J_         1         J_ 

X  ~  140       210  ~~  420' 

1     I     I     ,      *  J 

T  y  210  V  105 T  420      60 

hence,     -  =  105,      -  =  210;      -  =  420.         ,    *   ,      =  60. 

x  y  z  '     x  -f-  y  -f  z 

27.  A  has  3  purses,  each  containing  a  certain  sum  of  money.  If 
$20  be  taken  out  of  the  first  and  put  into  the  second,  it  will  contain 
four  times  as  much  as  remains  in  the  first.  If  $60  be  taken  from  the 
second  and  put  into  the  third,  then  this  will  contain  If  times  as 
much  as  there  remains  in  the  second.  Again,  if  $40  be  taken  from 
the  third  and  put  into  the  first,  then  the  third  will  contain  2J  times 
as  much  as  the  first.     What  were  the  contents  of  each  purse  ? 

Let    x    denote  the  number  of  dollars  in  the  first  purse. 
y         "  "  "  "      second  " 

z        "  "  "  "      third     " 

then  from  the  conditions  of  the  problem, 
4  (x  -20)  =  y  +  20, 
i  (y  -  60)  =  z  +  60, 
2_40    =  23  (* +  40); 

clearing  of  fractions,  performing  operations  and  transposing, 
4x-      y=    100     ....     (1) 
7y  -    Az  =    660  ...     (2) 

82  -23*  =  1240    ....     (3);      * 

combining     (1)     and     (2), 

28x-  4^  =  1360    ....     (4); 


102.]  EQUATIONS  OF  THE  FIRST  DEGREE.  57 

combining     (3)     and     (4), 

33*  =  3960;     .\     x  =  120  ; 

by  substitution,  y  =  380 ;     z  =  500. 

28.  A  banker  has  two  kinds  of  money ;  it  takes  a  pieces  of  the 
first  to  make  a  crown,  and  b  of  the  second  to  make  the  same  sum. 
Some  one  offers  him  a  crown  for  c  pieces.  How  many  of  each 
kind  must  the  banker  give  him  ? 

Since  it  takes  a  pieces  of  the  first  to  make  1  crown,  -  —  the  part 
of  a  crown  in  each  piece ;  and   -,    the  part  of  a  crown   in   each 

piece  of  the  second : 

let     x     denote  the  number  of  pieces  taken  of  the  first  kind, 
y  "  "  "  "  second  " 

from  the  conditions  of  the  problem, 
x  +  y  =  c 

x        II 

-  +  j-  =  1,     or     bx  -{-  ay  z=.  ab  \ 

by  combination,  % 

by  —  ay  =  be  —  ab;     or     y  (b  —  a)  =  b  (c  —  a) ; 

(a  —  c)  b  .  a  (c  —  b) 

.\     y  = f— ;      whence,     x  =  — -^. 

r         a  —  b  a  —  b 

29.  Find  what  each  of  three  persons,  A,  B,  C,  is  worth,  knowing, 
1st,  that  what  A  is  worth  added  to  /  times  what  B  and  C  are  worth, 
is  equal  to  p ;  2d,  that  what  B  is  worth  added  to  m  times  what  A 
and  C  are  worth,  is  equal  to  q ;  3d,  that  what  C  is  worth  added  to 
n  times  what  A  and  B  are  worth,  is  equal  to  r. 


58  KEY  TO  DAVIES*  BOURDON.      .  [102. 

Let    x    denote  what  A  is  worth, 
y         "         "      B         " 
z         "         "      C        " 
then,  from  the  expressed  conditions, 

x  +  l    (y  +  z)=p     .     .     .     .     (1) 
y  +  m(x  +z)  =  q     •     •     .     .     (2) 

9  4.  n  (x  +  y)  =  r     .     .     .     .     (3)  . 

wnich,  by  adding  and  subtracting  Ix,  my  and  nz,  may  be  written 

under  the  forms 

(1_  i)x+  i  (x  +  1/  +  z)=p  .     .     .  .     (4) 

(1  —  m)  y  -f  m  (x  -f  y  +  z)  =  q  •     •     •  •     (5) 

(1  —  ri)  z  +  n  (x  -f  y  -f  z)  —  r  •     •     •  •     (6)  ; 

dividing  both  members  of  each  equation  by  the  co-efficient  of  its 
first  term, 

•+it1  <«+*,+«)«  Hh   •  '  *  '  (T) 

y  +  r^-(*+y  +  2)  =  1-?—     ....    (8) 

1  —  m  v  '       1  —  m  x 

adding  these,  member  to  member,  and  deducing  from  the  resulting 
equation  the  value  of  x  -f-  y  +  ^ 

*   +  ,-*-  +   * 


1  —  /       1  —  m       1  —  n  .    • 

*+y+z= j '     '     (10). 

1  +  1 J+i +  1 

1  -—  /       J  —  m       1  —  n 

Denote  the  second  membei  of  equation   (10)   by  the  single  letter  st 


1 

p 

I 

Is 
1  — 

I 

q 

ms 

1 

m 

1  —  m 

r 

ns 

P- 

■Is 

1  - 

■I 

9- 

ms 

l  - 

m 

r  — 

ns 

102.]  EQUATIONS  OF  THE  FIRST  DEGREE.  59 

a  known  quantity.  Then  by  substituting  this  for  the  factor 
*  +■  V  +  zt  m  eacn  °f  tne  equations  (7),  (8)  and  (9),  and  deduc- 
>»g  the  values  of  ar,  y    and  z,  we  have, 

•  •  •  (ii) 

•    •    •    (12) 

!,_„        !_„!_„  •     *    *    (13>- 

Had  we  represented  the  polynomial  x  -j-  y  4*  i  5>J  s>  the  alge- 
braic work  would  be  slightly  diminished,  but  the  preceding  method 
has  been  followed  in  order  to  show  more  clearly  the  process  of  solu- 
tion. / 

30.  Find  the  values  of  the  estates  of  six  persons,  A,  B,  C,  D,  E, 
F,  from  the  following  conditions  :  1st.  The  sum  of  the  estates  of  A 
and  B  is  equal  to  a  ;  that  of  C  and  D  is  equal  to  b  ;  and  that  of  E 
and  F  is  equal  to  c.  2d.  The  estate  of  A  is  worth  m  times  that  of 
C ;  the  estate  of  D  is  wrorth  n  times  that  of  E,  and  the  estate  of  F 
is  worth  p  times  that  of  B. 

1st  Solution. 
Let     x     denote  the  value  of  A's  estate  ; 


a  —  x  « 

it 

"   B's       ' 

y 

a 

:'   C's       • 

•'-*•«■ 

a 

"   D's      ' 

z         " 

a 

"   E's      * 

c  —  z  '•  "      "   F's      " 

from  the  conditions  of  the  problem, 


60 


KEY  TO  DAVIES    BOURDON. 


[102. 


x  =  my 

x  —  my  =  0  •     •     • 

•  (i) 

b  —  y  =  nz 

-     or,      ■ 

y  +  nz  =  J  •     •     • 

•     (2) 

c  —  z  =  p  (a  —  x) 

j92r  —  z  =z  ap  —  c     •     • 

(3) 

combining  (1)  and   (2), 

x  4*  ffinz  =  bra     •     •     •     •     (4) ; 

combining  (3)  and  (4),  and  finding  the  value  of  z, 

bmp  +  c  —  ap 

which  denote  bv  P  • 

combining  (3)  and  (4),  and  finding  the  value  of  ar, 

amnp  -j-  bm  —  cmn         .  ,  ■ 

a:  = — ,  which  denote  by  Q\ 

mnp  4  1 

substituting  the  last  value  in  (1),  and  finding  the  value  of  y, 

anp  -\-  b  —  en 


mnp  -f-  1 


which  denote  by  R ; 


whence, 


A's  estate  is  equal  to         ft 


B's     " 

If 

«     a-  ft 

C's     " 

U 

ft 

D's    " 

u 

"     6-5, 

E's    « 

a 

i>, 

Fs     " 

It 

"     c  -  P. 

We  might  have  combined  (2)  and  (3),  eliminating  2,  and  then  from 
this  resulting  equation,  taken  with  (1),  have  found  the  value  of  ot 
and  y. 

2d.  Solution 
By  a  single  unknown  quantity, 
Let  x  denote  the  value  of  C's  estate ; 

then  will  b  -  X  "  "  D's       « 


116.]  INEQUALITIES,  61 

mx  denote  the  value  of  A's       " 

a  —  mx  "  "  B's        " 

p(a  —  mx)        "  "  Fs       " 

c  —  ^(a-ma:)"  "  E's       " 

and  from  the  remaining  condition    of  the  problem, 
b  «_  x  =  n \C<f~  p  (a  — •  ma;)]  ; 

whence,  by  the  rule  for  solving  equations  of  the  first  degree, 

b  -f-  anp  —  nc 

x  s= — -i — « 

mnp  -+•  1 

Having  found  the  value  of  C's  estate,  the  remaining  quantities 
may  be  found  by  substituting  it  in  the  expressions  of  the  data,  and 
reducing.     The  operations  are  obvious. 

INEQUALITIES. 

1.  Given,         5x  —  6  >  19,         to  find  the  smallest  limit  of  x< 
If  we  add  6  to  both  numbers  of  the  inequality,  we  have 

6#>19-f6,     or,     6ar>25; 
dividing  both  numbers  by  5,  we  have 

x  >5. 

14 

2.  Given,         Sx  -f-  — -  x  —  30  >  10,     to  find  the  least  limit  of  *. 

Reducing,  8x + ?*—  30  >  10,     or,     10;r  —  30  >  10  ; 

adding  30  to  both  members  of  the  inequality,  and  dividing  by  10, 
we  have, 

x>4. 

x      x      x       13       17 

3.  Given,         ^"~-o+o^"o^>"o"'  t0  **n<*  tne  *east  ^m^  °^  Xi 

\*  <L>  *»  -»  •* 


62  KEY  TO  DAVIES*  BOURDON.  L123. 

Multiplying  both  members  by  6,  we  have, 

x  —  2x  +  Zx  +  39  >  51; 

reducing,  subtracting  39  from  both  members,  and  dividing  by  2, 

we  have, 

#>6. 

OX  CL^ 

4.  Given,         — -  -+■  bx  —  ab  >  —  to  find  the  least  limit  of  x* 

5  5 

Multiplying  both  members  by  5,  we  have 

ax  -f-  56rc  —  5ab  >  a2 ; 

adding  -f  *>a6  to  both  members,  and  dividing  by  the  co-efficients  of 
x,  we  have, 

(a  -f  5b)  x  >  a  {a  -f  56) ;     or,     x  >  a. 

5.  Given,         — a#  -f-  a&  <  — ,  to  find  the  largest  limit  of  z> 

Multiplying  both  members  by  7,  adding  —  7ab,  and  dividing  by 
the  co-efficients  of  x,  we  have, 

(b  —  7a)  x  <  b  (b  —  7a)     or,     x  <  b> 
SQUARE    ROOT    OF    NUMBERS. 


ROOT. 

REMAINDERS. 

1. 

85. 

2. 

133. 

3. 

997. 

4. 

9256     . 

»•    .     .    5437. 

5. 

8234     .     j 

.     .     .  13919. 

6. 

1671     .     i 

,     .     .      160. 

7. 

6123     . 

.     .         4913. 

132-136.] 


SIMPLIFICATION  OF  RADICAL8. 


63 


60507. 
958510 


252G04. 


Note. — "Where  the  square  root  consists  of  six  or  more  places  of  figures,  a 
very  good  approximate  result  may  be  obtained  by  finding  by  the  rule  half  or 
more  than  half  of  the  figures,  and  then  having  formed  the  divisor  as  directed, 
bring  down  all  the  figures  and  proceed  as  in  simple  division  till  the  required 
number  of  places  of  figures  is  obtained. 

In  the  last  example,  having  found  the  figures  958,  we  have  a  remainder 
977  ;  bringing  down  the  remaining  figures  and  forming  the  divisor  as  directed, 
the  process  of  approximation  will  be  as  indicated 

1916)977672704(510...    . 


SQUARE  ROOT  OF  POLYNOMIAL 

ROOT. 

2.  a2  +  %ax  +  x2. 

3.  a2  —  ax  H-  x'2. 

4.  2z3  +  3z2  -  x  -f-  1. 

5.  3a2  -  2ab  +  4b2. 

6.  5a26  -  4abc  -f-  Mc2  —  Sate. 


SIMPLIFICATION  OF  RADICALS. 

EXAMPLES.  SIMPLIFIED. 


1.     ynbcMc. 


2.     y™W. 


3.     -v/32a968c. 


4.  -y/256a264c8. 

5.  V/1024a967c5. 

6.  ^/12SaP b*cH. 


5a-\/Sabc. 

ma?c^/Zb. 

4a4M-v/2ac^ 

16a6V. 

S2a*b3c2</aJc. 


2a362cV182aW- 


64  KEY  TO  DAVIES*  BOURDON.  1142-150 


TRANSFORMATION  OF  RADICALS. 


4.  Arts,  x-y/x  —a2. 

8.  Ans.  (2a  -f  24a;2)  y/b. 

9.  Ans.  191  y^ 

13.  Ans.  2-^/27 

14.  ji/w.  96  y/5a. 

EQUATIONS  OF  THE  SECOND  DEGREE. 

1.  Given        ■? —  -3=1- — a? :r>     to  find  the  values  of  x. 

So  ad 

Clearing  of  fractions,  transposing  and  reducing, 

a2  -  b2 


X2- 

;nce,  by  the  rule 

"sr*^ 

•T     2ab      ^V1^ 

—  2a2b2  -f  64 
4a262 

„2_^2        a8 +  6*. 
2a6     *     2a6     ' 

2a2       a 
taking  the  upper  sign,  x  =  ^  ==  -, 

2b2            b 
"        lower     "       x  =  —  5-j  as 

2a6  a 


(fo3^  1+c       x2 

~  +  t  +  l  -  — 


2.  Given h  —  +  *  =  — ! T  +  c? 


to  find  the  values  of  x. 

Clearing  of  fractions,  transposing  and  reducing, 

2  1  d2  ~  c         l 


150.1  EQUATIONS  OF  THE   SECOND  DEGEEE.  65 

whence,  by  the  rule, 
d2  —  c 


x  = 


2cd 


±      71      rf4-2«P  +  c8  <P  —  e      d2  +  c 

:Vc  4c2tf2  -  2cd      -     2crf    ' 


taking  the  upper  sign,         a;  =s  — -^  =  -, 

"         lower     "        a;  =  —  — -  = 

2cd  c 

0     „.  x2      2x  ,   59       0      x2       x          .    .    . 

3.  Given    — -  —  — -  -f  — -  =  8  —  — -  -  ^,    to  find  the  values  of  a?. 

Clearing  of  fractions,  transposing  and  reducing, 

2  2  5 

whence,  by  the  rule, 

l  ^     ATT     1 7  7 

ir  =  3±V4  +  9  =  3±6; 

hence,  a;  =  -      and     x  =  —  -• 

2  0 

*    <^>  90         90  27  _  a   . 

4.  Given — -  =  — — ,    to  find  the  values  of  a?„ 

x       x  +  1       a;  +  2 

Dividing  both  members  by  9, 

10         10  3 


x       x  +  1       x  +  2* 
clearing  of  fractions,  transposing  and  reducing, 


2          7             20 
Xz X  = • 

3  3' 


whence,  by  the  rule, 


66  KEY  TO  DAVIES*  BOURDON.  [150. 


7  /20       49       7       17 


hence,  x  as  4,     and     *  a*  —  —• «  —  5    . 

b  o 

5.  Given         — — 2  sb £-,  to  find  the  valuei  of  ar. 

8  —  a:  x  —  2 


Clearing  of  fractions,  &c, 


„       39  28 


whence  by  the  rule, 


39  _,      A~28 

io±V-T 


1521  _  39      31 

+   100  ~~  10       10 


hence,  x  =  7,     and     rr  =  —  =  -. 


x2  6  —  1  6 

•6.  Given        or —  -+-  b  ss — = —  x2  -f-  -#?  to  find  the  values  ofx. 

o  o  a 

Clearing  of  fractions,  &c, 

2       a2-b  A 

x1 x  =  o  ; 

a 
whence,  by  the  rule, 


a*  -  b  L       a4  —  2a26  +  &2       a2  —  5       a2  -f& 

sr  *%/*  +  — s? — ""t^-ir1 


hence,  x  =  a,         a;  = 


&—  &      ,  3*3       a2       6  +  «      ,   «"       &2 

c 
to  find  the  values  of  ar, 


7.  Glven         ___,+  «,+ 


161.]  EQUATIONS  OE  THE  SECOND  DEGREE*  67 

Clearing  of  fractions)  &c., 


„       2b         a2  -  £2 
xz x  £z 


whence)  by  the  rule, 


6  / 


c2  c2  ~~  c       c 


,                                      6  -f-  «  ,  6  —  a 

hence)  x  az —,    ana    * 


C     '  c 

8.  Given         mx*  -f  mti  s=  2my/n  •  a;  -f  *we2,  to  find  the  values  of  ts 

Transposing  and  reducing, 

2m  Jn~  mn 

*2 y—  x  s=  — ; 

to  — ■  n  to  —  ft 

whence,  by  the  rule, 
to 


x  we,  — 
to 


y^  /_     mn  m*?t         ^  m\/5t        n^/m 

—  /i      V       to  —  n       (to  —  lip  ~~  to  — -  n      to  —  n  ■ 


m*/w  ■+-  n\fm            \/mn  ( -i/ro~-f-  a/w)  ^/frm 

hence    £  ssc  — — — ~ —  —  r    y      *__       j 

since,         to  —  rt  s=  (-y/ro  +  y^n)  (^/m  — ■  -vAF        (Art.  47.) 
also,, 

^_  to  yn~-  n\/m  ^  yrnn  (ym—  -\/ft)  yWi 

»»  —  ft  (V^+  -\A)  (V^—  V^)      V^+ V* 

2c£  Solution. 
mx2  -f  tow.  =  2to  y^tT-  a;  -f  wz2 ; 
transposing  the  first  term  of  the  second  member,  we  have» 


68  KEY  TO  DAVIES*  BOURDON*  [15L 

mx2  —  2m  yn  •  x  +  mn  ss  nx2  5 
observing  that  the  first  member  is  the  square  of  a  binomial  whose 

terms  are     \/m  •  a:  —  ^/m  -yfn^   we  have, 

•y/m  •  x  —  -y/m  -y/n  a=  dfc  *y/7i  *  a:; 
.t*»,  yw  •  a;  ^=  yV«  x  =  ym  yn 

(^/rrJT^P  -y/n)  x  s=  ^/m^/n  :  hence, 

s/mn  .  i« 


and,    x 


y?/i  —  -y/rc  -y/m  4-  -/» 


,  „       6a2,    h2x       ah  —  262       3a2 

9.  Given        abx2 4 = ; x1 

c2         c  cl  c 

to  find  the  values  of  x. 

Clearing  of  fractions,  &c, 

2      b2  +  3a2     _  ah  —  2&2  +  6a2 
*   +      a*tf      P"  a**2  "' 

whence,  by  the  rule, 

h\/- 


2aJc~        V  a6c2  +  4a%2c2 


g  -f  3a2       3a2  +  4aS  -  &2  ■ 
*  2a6c  2afor  ' 

4a6  —  2£2      2a  —  0  6a2  -f  4a5  3a  +  26 

hence,   x  s  — ^— ; = »    a;  ss ^7-7 sa = • 

2abc  ac  2abc  be 

,,     ~*  4a:2    ,   2a:  ,    mA       mg.       3a:2  ,   58a: 

10.  Given         —  -f  y  +  10  =  19  -  y  4-  -y* 

<»  find  the  values  of  x. 


161 J  EQUATIONS  OF  THE  SECOND  DEGREE. 

Clearing  of  fractions,  &c., 

z2  —  Sx  =  9  ; 
whence,  by  the  rule, 


*  =  4rfcv/9-flt>  =  4=fc5j 
hence>  a;  =  9,     x  =  —  1. 

11.  Given        l±-a__  *  =  ?_=:.* 
x  —  a 

Clearing  of  fractions,  &c., 


x ,    to  find  the  values  of  x. 


(M-2)a^ 

r  6-2  ■ 

whence,  by  extracting  the  square  root  of  both  members, 


-**tflS  -  — -v/?? 


12.  Given        2*  +  2  =  24  -  5s  -  2**,  to  find  the  values  of  * 
Transposing  and  reducing, 


*2-f  gr  =  ll 


whence,  by  the  rule, 


*  ?±v/11  +  49  7       15. 

4       V       T16  4       4' 

whence,  *  =  2,     and    «  =  -  11. 

2 

13.  Given        *»  _  «  _  40  ==  170,     to  find  the  values  of  *. 
Transposing, 

z2-x=z  210 : 


70  KEY  TO  DAVIEs'  BOURDON.  [151. 

whence,  by  the  rule, 


hence,  *  =  15,    and    x  =  —-14. 

14.  Given        3s2  -f  2x  —  9  =  76,    to  find  the  values  of  x. 
Transposing  and  reducing, 


2         85 
whence,  by  the  rule, 


*2  +  3*-3.> 


1  /85   ,    1  1       16 

17  2 

hence,  x  =  5,     and    x  =  — —  =  —  5  -• 

3  3 


15.  Given        a2  -f  b2  —  25a?  +  x2  =  — — ,  to  find  the  values  of  x. 

n2 

Clearing  of  fractions,  &c, 

2i      2bn2       -  n2a2  +  n2f)2 . 

w2  —  n2    "      m2  —  n2    ' 

whence,  by  the  rule, 
bn2 


nv 


r  /nW+_nW  b2n* 

■n2  ~  V    m2  —  n2  m*  —  2m2w2  +  w* 


bn2  n  y/  a2m2  +  b2m2  —  a2n2 

n2  —  m2~  n2  —  m2 


whence,         x  =  n2  __  m2   \  bn  ±  y/a2m2  +  b2m2  —  a2n2  [ 


154.1  EQUATIONS  OF  THE   SECOND  DEGREE.  71 


PROBLEMS  GIVING  RISE  TO  EQUATIONS  OF  THE  SECOND  DEGREE. 

4.  A  grazier  bought  as  many  sheep  as  cost  him  £60,  and  after 
reserving  15  out  of  the  number,  he  sold  the  remainder  for  £54,  and 
gained  2s.  a  head  on  those  he  sold  :  how  many  did  he  buy  ? 

Let    x     denote  the  number  purchased  : 
and  x  —  15,     the  number  sold  ; 

then  will      denote  the  number  of  shillings  paid  for  1  sheep, 

and  —      the  number  of  shillings  received  for  each. 

x  — — 1«> 


From  the  conditions  of  the  problem, 


1200         1080 
x      ~s-15          ' 

clearing  of  fractions,  &c, 

x*  +  45*  =  9000  : 

whence,  by  the  rule, 

*  =  -425±v/9ooo  +  2(f  =- 

45       195 

2         2    ; 

hence.  x  =  75,     and     x  =  —  120, 

the  positive  value  only,  corresponds  to  the  required  solution. 

5.  A  merchant  bought  cloth  for  which  he  paid  £33  15s.,  which  he 
sold  again  at  £2  8s.  per  piece,  and  gained  by  the  bargain  as  much 
as  one  piece  cost  him  :  how  many  pieces  did  he  buy  ? 

Let     x     denote  the  number  of  pieces  purchased  : 

675 

then  will,         — ■     denote  the  number  of  shillings  paid  fbr  each, 


72  KEY  TO  DAVIES*  BOURDON.  154. 

and  48«  the  number  of  shillings  for  which  he  sold  the  whole. 
From  the  conditions  of  the  problem, 

48z-675=  — ; 

then,  by  clearing  of  fractions,  &c, 

„      225         225 

xl x  —  • 

16  16 

whence,  by  the  rule, 


225  /  225       50625       225      255 

32  ~  V    16  +   1024    ~  32  ±  32  ; 


.       ,  •  •  480       ■ 

using  the  positive  value  only,         x  =  — —  =  15. 

6.  What  number  is  that,  which  being  divided  by  the  product  of 
its  digits,  the  quotient  will  be  3 ;  and  if  18  be  added  to  it,  the  order 
of  its  digits  will  be  reversed  ? 

Let    x    denote  the  first  digit, 
and,       y        u  second  " 

then  will  10#  -f  y        denote  the  number. 

From  the  conditions  of  the  problem, 

10s  +  2/+18  =  10y  +  *; 
whence,  by  reduction,  * 

10a;  -f  y  =  3xy, 

y-x  =  2-, 

finding  the  value  of  x  in  terms  of  y  from  the  second,  and  sub* 
stituting  in  the  first,  we  have, 


154.]  EQUATIONS   OF   THE    SECOND   DEGREE.  73 

whence,  by  transposing,  &c., 


17  „  20 

by  the  rule, 


y2~yy2  =  --3-'    and' 


17  /      20   ,   289        ,    17       7 

7*  V  "T+ir'+T** 


taking  the  positive  sign,  y  =  4  ; 

whence,  x  22  2,     and  the  number  is  24. 

7.  Find  a  number  such  that  if  you  subtract  it  from  10,  and  mul 
tiply  the  remainder  by  the  number  itself,  the  product  will  be  21. 

Let    x    denote  the  number : 

from  the  conditions  of  the  problem, 

(10  -  x)  x  =  21 ;     or,     x2  —  10*  =  -  21 ; 

by  the  rule, 

z  =  5±,v/-21+25=:5:fc2; 

whence,  \       x  =  7,     and    x  —  3. 

8.  Two  persons,  A  and  B,  departed  from  different  places  at  the 
same  time,  and  travelled  towards  each  other.  On  meeting,  it  ap- 
peared that  A  had  travelled  18  miles  more  than  B;  and  that  A 
could  have  performed  B's  journey  in  15J  days,  but  B  would  have 
been  28  days  in  performing  A's  journey.     How  far  did  each  travel? 

Let       x        denote  the  number  of  miles  B  travelled ; 
x  +  18         "  "  "     A         " 

l|l  "  "  "     A         "        in  one  day; 


t:  B 


74  KEY  TO  DAVIES'  BOUKDON.  [154"55. 

—X- —    denote  the  number  of  miles  B  travelled  in  one  day  , 

*/+18v       "  "  days  A 

\  15}  / 

*  « 

from  the  conditions  of  the  problem, 

x+  18  _        x_  x2  -I-  36a;  +  324  _  x2   # 

__x__    ~  aT+ .18         °r'  28~~         "  15f 5 

15f  28 

clearing  of  fractions,  and  reducing, 

,       324         2916 

X1 X  =  • 

7  7 


By  the  rule, 

162         /2916 

*=    7    -V     7 

26244 
+     49 

162      216 
7           7    ' 

hence,  using  the  upper  sign, 

<r  =  -— -  =54. 

7 

9.  A  company  at  a  tavern  had  £8  15s.  to  pay  for  their  reckoning ; 
but  before  the  bill  was  settled,  two  of  them  left  the  room,  and  then 
those  whc  iemained  had  10s.  apiece  more  to  pay  than  before:  how 
many  were  there  in  the  company  ? 

Let  x  denote  the  number  in  the  company. 


155.]  EQUATIONS  OF  THE  SFCOND  DEGREE.  75 

175 

Then      —   will  denote  the  number  of  shillings  each  should  pay; 

X 

175 

-  "  '♦  «  "    paid; 


x 

from  the  conditions  of  the  problem, 
175         175 


10; 


x  —  2  re 

clearing  of  fractions, 

175*  -  175*  +  350  =  10s2  -  20*; 
whence,  x2  —  2*  =  35. 

By  the  rule, 

*  =  1  ±^/5G=  1  rb  6; 

using  the  upper  sign,         *  =  7. 

10.  What  two  numbers  are  those  whose  difference  is  15,  and  o* 
which  the  cube  of  the  lesser  is  equal  to  half  their  product? 

Let       *       denote  the  smaller  number; 
then  will  x  +  15  "  greater         " 

from  the  conditions  of  the  problem, 

x*  =  ±{z*  +  lbx),     or,     *2  =  i(*  +  15); 

2       *         15 
whence,  xl  —  -x  =— ■ . 

By  the  rule, 


1  /15        1        1       11 

4  V  2    M6       4~  4  ' 

using  the  upper  sign,  x  —  3  ;         hence,         *  -f  15  =  18. 

11,  Two  partners,  A  and  B,  gained  $140  in  trade:  A's  money 


76 


KEY  TO  DAVIES'  BOURDON.  [155. 


was  3  months  in  trade,  and  his  gain  was  $60  less  than  his  stock :  B's 
money  was  $50  more  than  A's,  and  was  in  trade  5  months:  what 
was  A's  stock  ? 

I-<et         x  denote  the  number  of  dollars  in  A's  stock ; 

x  -f  50  "  "  "  B's    " 

x  —  60  "     A's  total  gain  ; 

"     A's  gain  per  month  ; 


3 

-60 


Sx 


"     A's     "        "        «    on  1  dollar ; 


(— -^)(z  +  50)   B's    "        "    .     « 
(— ^— j  (x  +  50)  5  B's  total  gain. 


From  the  conditions  of  the  problem, 


s_60  +  (*  5  6°;  (x  +  50)  5  =  140; 

oX 


clearing  of  fractions,  and  reducing, 


4 


By  the  rule, 


325         /,    10^c    ,   105625      325      475 
*  =  — ±N/+1875+nf _=_=fc_; 

whence,  x  =  100. 

12.  Two  persons,  A  and  B,  start  from  two  different  points,  and 
travel  toward  each  other.  When  they  meet,  it  appears  that  A  has 
travelled  30  miles  more  than  B.  It  also  appears  that  it  will  take  A 
4  days  to  travel  the  road  that  B  had  come,  and  B  9  days  to  travel 


158.]  EQUATIONS  OF  THE  SECOND  DEGREE.  77 

the  road  which  A  nad  come.     What  was  their  distance  apart  when 
they  set  out? 

Let         x  denote  the  number  of  miles  B  travelled; 

then  will    x  +  ZO    "  /  "  "A 


4 

*  +  30    k 
9 

*+_30    ti 

S 

(NN" 

From  the  conditions, 


"      A  travels  per  day ; 

it  g  (i  M 

days  A        " 

» 


x  x  +  30  xz       x2  +  60*  +  900 

or,     — -  =  — 


9 


m  §   ' 4 

whence,  by  reduction, 

'  a;2-~4&c=s720' 
and  by  the  rule, 

x  *  24  db  V72CTf576  -  24  =fc  36 ; 
taking  the  upper  sign,         x  =a  60.     and     £  -f  30  =2  90 ; 
hence,  the  distance  is  150  miles. 

EXAMPLES  INVOLVING  RADICALS  OF  THE  SECOND  DEGREE. 


n  I  q2   /j»2  /j« 

3.  Given         -  +  \  / ~ —  =  t>    to  find  the  values  of  x< 

x        \       x2  b 


78  KE?  TO  DA  VIES*  BOtJfcDoN.  [166. 

Multiplying  botn  members  by  bx,  and  transposing, 


b  ya2  — ■  x1  =z  x2  —  ab  ; 
squaring  both  members, 

b2a2  —  b2x2  -.it  x*  -  »  2afo2  -f  a2b2 ; 
cancelling  62a2,  dividing  both  members  by  x2  and  transposing, 
<K2  =  2a5  -  b2         k  \         a;  ==  ^h  -y/Zab  —~b2. 

4.  Given       * /i±?  +  2  \/— £- =  b2  x/—?—,    to  find  « 
Multiplying  both  members  by  a/ — -> 

a;  V    a: 

multiplying  both  members  by  x,  and  transposing, 

2^/ax  sb  62x  — -  re  — ■  a  s=  (b2  —  1)«  —  a  5 
squaring  both  members, 

4a<r  =  (M  -  262  +  1)  «*  -  2a  (jM  r-l)i  +  a2; 
transposing  and  reducing, 

2_     2<b*  +  *)      fe        ^2_ 

*        6*  -  262  -r-  I*"  ^-262+l* 

Whence, 


q(&2  -f  1)  /_  »2  qg(6*  +  262  +  1) 

*  m  64  _  262  +  1        V       F^-  262  -f  1        (&4  -  262  +  1?' 

«(62-fl)  2ao 

P  -  262  +  1   _  64  -  262  -f  1 ' 

now,  6*  ~  262  4-  1  «=  (*  -  1)2(6  +  1)\ 


166-7.]  EQUATIONS  OF  THE   SECOND  DEGREE.  79 

Hence,  taking  the  upper  sign  and  reducing^ 

_  q(6-fl)2  _         a(b  +  l)2  a 

X  "   (62  -.  I)5  ~  (6-  1)2(6  +  l)2  "  (6  -I)5 ; 

and  taking  the  lower  sign  and  reducing, 

_  a(b  -  l)2  _         a(b  —  l)2  a 

x  ~  (p  ~_  1)2  ~  (6  ^-  1)2(6  +  l)2  *  (6  -f  l)8' 

or,  uniting  the  two  values  in  a  single  expression, 

a 


(4  qz  i)« 


5.  Given,         —.       ■    .  sc  6,     to  find  £. 

a  -f  y^  a2  —  a2 

Clearing  of  fractions,  transposing,  &c, 

a(l  -  6)  &  (6  +  1)  Va2  -  a;2 } 

squaring  both  members, 

a2 (1  -  6)2  =  (6  +  l)2 (a2  -  z2) 5 

transposing  and  reducing, 


4«2& 

'  (*  +  I)2 

2a  Ji 

s/x  -{-  V^  ~~  a 

,     to  find  x. 

6.  Given, 

■\/x  —  -y/x  ~-  a       x  —  a 


Multiplying  both  terms  of  the  first  member  by    ifx  -*  y/x  -«*  a% 
and  then  dividing  both  members  by    a, 

1  __     0 

2x  —  a  —  2-v/a2  —  a#       #  ■*-  a 

clearing  of  fractions,  &c, 


80  KEY  TO  DAVIES'  BOTTKDON.  [107. 

(1  —  2n2)z  —  (1  —  n2)a  =  —  2n2  y^2  —  ax  ; 

squaring  both  members,  transposing  and  reducing, 

2  (1  -  3ti2)  a  1  —  2>i2  +  n* 

l-4n2  1  -  4n2  ' 

whence  by  the  rule, 


1  -  3n»  /      (l-2n2+rc*)a2         (1  -  6tt2  +  9m*) 

1  _  4^2 a       V  1  _  4w2  +         (1  -  4rc2V2~ 


(1  —  3w2)  2rc3a      _  (1  -  3/i2  dfc  2«3)  a 


1  _  4n2  1  -  4w2  1  —  4m2 

Taking  the  upper  sign,  and  dividing  both  terms  of  the  fraction  by 

1  +2n, 

_  (1  —  2n  +  rv*)a,  __  (1  —  nfa_ 
X  **         P^  "    .  1  -  2n~ ' 

Taking  the  lower  sign,  and  dividing  both  terms  by  1  —  2n, 

—  g(1  +  jfo  +  n2)     ,.  (1  +  nfa 

X~~  1  +  2n  ~     1  +  2/i    ' 

«0  ±w)2 
taking  the  two  values  together,  x  as      '  •     • 


7.  Given  Y — =  +     — ; =\/  t>  to  "na  *• 

Jx  Jx  V  5 

Multiplying  both  members  by  y^ 

squaring  both  members, 

2a  +  2ya2-a:2=:y;     0r,     2</«2  -  *2  =  J  ~-2a' 


167.]  EQUATIONS  OF  THE  SECOND  DEGREE.  81 

squaring  both  members, 

cancelling  4a2  in  the  two  members,  and  dividing  both  by  re2, 

_  x2      4a 

clearing  of  fractions,  &c, 


x2  =  Aab  —  462  .  • .  x  =  ±  2  Jab 


ft* 


a  -f  re  +  <\/2az  +  a;2       .  -, 

8.  Given ?- =  b  :  to  find  #. 

o  +  * 

Clearing  of  fractions,  transposing  and  factoring, 


<y/'2ax  +  x2  =  (b  —  1)  (a  +  x). 
Squaring  both  members, 

2ax  +  x2  =  (b2  -26  +  1)  (a2  +  2ax  +  x2) ;     or, 
2ax  +  z2  =  {62  —  26)  (a2  -f  2as  +  a;2)  +  a2  -f-  2ax  +  s2; 
whence,  by  reduction, 

whence,  by  the  rule, 


/       .  /I  -  26  +  b2\    ,     ;  1 

,=  -a±^«2(-^^^)  +  a2=-.±«^^2; 

1.        t                                     a(l--/2&~-ft*)  * 

taking  the  upper  sign,       z  = ^         ^  • 

♦  w      a    i  •  afl+v^"^) 

taking  the  lower  sign,       x  = J ~-         — - ; 

y/'lb  —  62 

^a(l^y/26-62 


whence,  *  =  ± 


/26-62 


82  KEY  TO  DAVIES'  BOURDON.  [167-69. 

2d  Solution. 
Make  x  +  a  s=  y ;  whence,  2ax  -f  xz  =  y2  —  a2 

substituting  in  the  equation,  and  clearing  of  fractions, 

V+  vV  —  a2  =  by  ; 
transposing,  &c., 


squaring  both  members,  and  cancelling, 

-  a2  =  (62  -  26)  y2 ; 
■whence,  solving  with  respect  to  y, 

a 

y=  ± 


J2b  -  b2 
substituting  for  y  its  value,  &c, 


a 

x  —  —  a  d= 


y^U' 


wlience,  as  before,         r  =_  - 


^26  -  P 

TKINOMIAL   EQUATIONS. 

3.  Given        ^*  -  (26c  +  4a2)  x2  =  —  62c2  ;  to  fine?  *. 
By  the  rule, 


c  =  =fc  y'&c  +  2a2  ±  ^/4a26c  +  4a*  =  ±y'6c  -f  2a2  ±  2a<v/fo~-f-  <S 

4.  Given        2*  —  7  yfx  =  99  ;     or,     2x  -  99  =  7  ^ 
Squaring  both  members 

4*2  -  396*  +  9801  =  49* ; . 


169-72.]  EQUATIONS  OF  THE  SECOND  DEGREE.  83 

transposing  and  reducing, 


445 

9801 

X2 

—  V.    ^—  — 

4 

4 

q  rule, 

445 

v- 

9801 
4 

,  198025 
+   64  > 

445 

203 

X 

»— • 

£  — — 

• 

8 

8 

or, 


taking  the  upper  sign,         x  t±  —  =  81 

242       121 

"       lower  sign,         x  =  — -  =  -j- 


(X  c 

5.  Given,  -  —  bx4  -f  -  x2  =  0,     to  find  x  \ 

transposing  and  re'ducing. 


whence,  «»  ±  Vgj^^p+-|^» 


reducing,  *  a  =fc  ^/  c  j  V*^"± 


EXAMPLES  OF  REDUCTION  OF  EXPRESSIONS  OF  THE  FORM  OF 


y«  ±  Vk 


Reduce  to  its  simplest  form,   v/28  +  10  V3. 
a  =  28,     6  *  300,     c  as  22. 


84  KEY  TO  DAVIES*  BOURDON.  [172. 

Applying  the  formula  and  considering  only  the  upper  sign, 


y/28  +  10^3  =  5+^ 


5.  Reduce  to  its  simplest  form,    wl  -f  4-y/—  3. 

a  as  1,         6  as  —  48,        c  =  7 ; 
applying  the  formula,  &c, 


*/l  -f  4V-T  =a  2  +  </=!£ 
6.  Reduce  to  its  si  molest  form, 


yjhc  +  26^/6^  62  -  \AC  -  26 Vbc  ~  R 

a  =  6c,         4  as  462  (6c  —  62),         c  =  6  (c  -  26)  5 
applying  the  formula  to  the  1st  radical, 


W6c  -f  26^57"^-  62  =  dfe  (yK  -  62  +  6) ; 
applying  the  formula  to  the  2d  radical, 


y/bc-Zby/biT^b'2^  ±(^-63-6); 
subtracting  the  second  result  from  the  first, 


^/M-  26  y^  -  62  -,/fc  -  26  V6C  -  62  =  ±  2*. 
7.  Redupe  to  its  simplest  form, 


yjab  +  4c2  -  fc  —  2^/4abc2  —  abd*: 
a  =  a6  +  4c2  —  d2,       6  ==  16a6c2  -  4a6^,       c  =  ab  —  4^  -f  d2. 


178.]  EQUATIONS   OF   THE   SECOND   DEGREE.  85 


t 


Applying  the  formula, 


<J ah  -f  4c2  —  <£*  -  2  JAabo*  —  abd*  =  ±  (yfib  -  y4c2  —  d2). 


EQUATIONS  OF  A  HIGHER  DEGREE  THAN  THE  FIRST,  INVOLVING 
MORE  THAN  ONE  UNKNOWN  QUANTITY. 

.  „•  ix2y  +*v2  =  6-  •  •  0)) 

5.  Given,  -J  "  .  _h    to  find  a;  and  y. 

tsy +  «y  =  12...  (2)1  y 

Dividing  (2)  by  (1),  member  by  member, 

o  2 

xy  —  2,     or    a;  =  — 

Substituting  this  value  of  a;  in  (1)  and  reducing, 

clearing  of  fractions  and  reducing, 

y2_3y  =  -2: 


l  3  /     ~   ,    9        3        1 

whence,  y  =  -  ±^ -2  +  ~  =  -  ± -; 

or,  y  =  2,     and     ys:l; 

whence  »=1,     and     a:  =  2. 

a:2  +  *  +  y  =  18  -  y2  .  .  ..  (1)  ) 
xy 


6.  Given,      ■<  ,  v      to  find  ar  and  y. 

^ (2)  j  y 


Multiplying  both  members  of  (2)  by  2  and  adding  the  resulting 
equation  to  (1),  member  to  member, 

X2  +  2xy  +  y2  +  x  +  y  =  30, 

or,  (a:   +  y)2  +  *  -f  y  =  30  ; 


UHI7BRSIT 


86  KEY  TO  DAVIES'  BOURDON.  [179. 

whence,  by  the  rule, 

whence,  x  -f  y  =  5,     and    x  +  y  =  —  6. 

Taking  the  first  value  of  a;  +  y  and  substituting  in  it,  for  y  its 
value  —  derived  from  (2), 

a; 
clearing  of  fractions  and  reducing, 

a;2  —  5x  =  —  6  ; 


5  /  25       5        1 

whence,  .  =  f  ±y/-  6  +  T  =  -  ±  -; 

vr,         a:  =  3,     and     x  =  2 ;     whence,     y  =  2,     and    y  as  3. 
Taking  the  second  value  of  #  +  y  and  proceeding  as  before, 

clearing  of  fractions,  &c,  , 

x2  +  6#  =  —  6 ; 
whence,  *  =  —  3  =fc  y/—  6  +  9  =  -3  +  /^"j 

and  by  substitution,  y  =  —  3  ^-y/3. 

PROBLEMS   GIVING  RISE  TO   EQUATIONS  OF  A  HIGHER  DEGREE 
THAN  THE   FIRST  CONTAINING  MORE  THAN  ONE  UNKNOWN 
•     QUANTITY. 

2.  To  find  four  numbers,  such  that  the  sum  of  the  first  and  fourth 
shall  be  equal  to  2s.,  the  sum  of  the  second  and  third  equal  to  2/, 
the  sum  of  their  squares  equal  to  4c2,  and  the  product  of  the  first 
*nd  fourth  equal  to  the  product  of  the  second  and  third.        m 


180.] 

Assue 

ning  the 

EQUATIONS 
equations, 

OF  THE 

SECOND 

DEGREE. 

» 

u  +  z 

=  25      . 

«     •      » 

(i) 

x  +  y 

=  2*'    • 

.     .     . 

(2) 

u2  +  x* 

'  +  y2  +  z2 

=  4c2    • 

.     .     . 

(3) 

uz 

=  ary     • 

.     .     . 

(4) 

87 


Multiplying  both  members  of  (4)  by  2,  and  subtracting  from  (3), 
member  from  member, 

u2  —  2uz  +  z2  +  x2  +  2xy  +  y2  =  4c2 ; 

or,  (w  —  z)2  +  (%  4-  y)2  =  4c2. 

Substituting  for  a;  +  y  its  value  2s'  and  transposing, 

(u  -  z)2  =  4c2  -  4s'2, 

or,  w  —   0  =  -/4c2  —  4s'2. 

Combining  with  (I), 

y4c2  _  45'2 


w  =  «  +  - = ;  =  s  -f  Vc2  —  s'2> 


^4C2  _  4s'2  

and,  s  =  s  —  2 ^ =  *  —  V°2  —  *'2  i 

reversing  the  order  of  the  members  of  (4),  and  proceeding  as  before, 
we  find  in  like  manner. 


x  =  s'  +yc 


and,      •  y  —  s'  —  -y/c2  —  s2. 

4.  The  sum  of  the  squares  of  two  numbers  is  expressed  by  a, 
and  the  difference  of  their  squares  by  b  :  what  are  the  numbers  ? 

Let  x  and  y  denote  the  numbers. 

From  the  conditions  of  the  problem, 


88                                       KEY  TO  DA  VIES' 

BOUKDON. 

x2  -f  y2  =  c   .• 

x2  -  y2  =  b     . 
By  adding,  member  to  member, 

•  •  •  (1) 

•    -    •     (2)-_ 

2x2=za  +  b 
by  subtracting, 

.           4-        /°  +  *. 

2y2  =  a  —  b 

/a -6 

y  =  ±v-ir- 

[181. 


5.  What  three  numbers  are  they,  which,  multiplied  two  and  two, 
and  each  product  divided  by  the  third  number,  give  the  quotients; 
a,  b,  c  1 

Let  x,  y  and  s,  denote  the  numbers 
From  the  conditions  of  the  problem, 

—  =  a    or,     xy  =  az     •     •     •     (1) 

■^  =  b     or,     yz  =  bx     •     •     •     (2) 

xz  fO\ 

—  =  c     or,     xz  =:  cy    •     •     •     (3). 

Multiplying  (1),  (2)  and  (3)  together,  member  by  member, 

x2y'zz2  __  alCxyz ; 
dividing  both  memoers  by  xyz, 

xyz  =  abc     •     *     •     (4) ; 

substituting  in  (4)  the  value  of  xy  taken  from  (1),  and  dividing  both 
members  by  a, 

z2  =  be  .  • .  z  =  V6c. 

Substituting  the  value  of  yz  and  dividing  by  b, 


182.]  EQUATIONS  OF  THE  SECOND  DEGREE.  89 

x2  =  ac  .  * .  x  =  Jac. 

Substituting  the  value  of  xz  and  dividing  both  members  by  c, 
y2  =  ab  .*.  y  —Jab. 


6.  The  sum  of  two  numbers  is  8,  and  the  sum  of  their  cubes  is 
152 ;  what  are  the  numbers  % 

Let  x  and  y  denote  the  numbers. 

From  the  conditions, 

*  +  y=    s (i) 

*3  +  2,3  =  152 (2); 

cubing  both  members  of  (1), 

x*  +  Sx2y  +  Zxy2  +  y2  =  512     •     •     .     •     (3) ; 

subtracting  (2)  from  (3),  member  from  member,  and  dividing  both 
members  by   3, 

x2y  +  xy2  m  120     ...     .     (4)  ; 
substituting  the  value  of  a;  taken  from  (1), 

(64  -  Uy  +  y2)  y+(8-y)y2  =  120, 
or,  reducing,  y2  —  8y  —  —  15 ; 


whence,      y  =  4  ±  -y/  —  15  -|-  16  =  4  ±  1     .  • .     y  =  5,     y  =  3 ; 
whence,  from  (1)  x  =  3,     a;  =  5. 

7.  Find  two  numbers,  whose  difference  added  to  the  difference  ot 
their  squares  is  150,  and  whose  sum  added  to  the  sum  of  their 
squares,  is  330. 

Let  x  and  y  denote  the  numbers. 

From  the  conditions  of  the  problem, 


90  KEY  TO  DAVIES'  BOURDON.  [182. 

xz  —  y2  +  x  -  y  =  150     •     •     ■     .     (1), 
x2  +  y2  +  x  +  y  =  S30     •     .     •     •     (2); 
adding  member  to  member,  and  reducing, 
x2  +  x  =  240 ; 


whence,  x  =  —  -  db  y240  -f  -  —  —  -  ±  — - ; 

or,  considering  only  the  positive  solution, 

x  =  15; 
whence,  from   (1),  by  substitution, 

y  =  9. 

8.  There  are  two  numbers  whose  difference  is  15,  and  half  their 
product  is  equal  to  the  cube  of  the  lesser  number :  what  are  the 
numbers  % 

Let  x  and  y  denote  the  numbers. 

From  the  conditions  of  the  problem, 


x  —  y  = 

:  15 

•     • 

• 

(1). 

xy 
2 

y\ 

or     x  = 

2y2 

•     • 

• 

(2); 

substitul 

:ing 

in(l) 

an( 

I  dividing 

both  members 

by  2, 

whence, 

• 

y  = 

1 
:4 

2    y 

V        2~ 

15 
2  ' 

1 

n 

4 

-Jfr 

1 
16" 

•• 

considering  only  the  positive  solution, 

y  =  3  ;     whence,  from  (1),     x  =  18. 


182.]  EQUATIONS  OF  THE  SECOND  DEGREE.  91 

9.  What  two  numbers  are  those  whose  sum  multiplied  by  the 
greater,  is  equal  to  77 ;  and  whose  difference,  multiplied  by  the 
lesser,  is  equal  to  12? 

Let  x  and  y  denote  the  numbers. 

From  the  conditions, 

(x  +  y)  x  z=  77,     or    *»  +  zy  c=  «H     •     •     •     (1); 

(x-y)y=\2,     or     xy  -  y*  =  12     •     •     •     (2); 

make     x  =  py;     whence, 


(^+^  =  77,       orj       y2  =  ^_       .       .       .       (3)> 


77 


(p   -1)^  =  12,     or,     V2=y~    ...     (4); 
equating  the  second  members  and  reducing, 

f   -12^=-12' 


whence,        jr«^±y^-T*4 


77      4225  _  65      23 
12.       576  ~  24*24' 

taking  the  upper  sign, 

_88_  11^ 
P  ~  24  ~  3  ; 

/Of*  O 

substituting  it   (4),         y  =  y  -g-  =  g-  V^J 

whence,  a;  =  —  -y/2; 

42      21 

taking  the  lower  sign,  p  =  —  =  — ; 


92  KEY  TO  DAVIES'  BOURDON.  [182. 

substituting  in  (4), 

y=z\/-— -  =  4;       whence,      *  =  7. 

10.  Divide  100  into  two  such  parts,  that  the  sum  of  the!*  squar 
roots  may  be  14. 

Let  x  and  y  denote  the  parts. 
From  the  conditions, 

*  +      y  =  100     .     •     •     (1), 

y^+  -v/y=    14     ...     (2): 

squaring  both  members  of  (2)  and  subtracting  (1),  member  from 
member. 

2  i/xy  =  96,     or     ^/xy  ==  48,     or     xy  =  2304 : 
substituting  for  y  its  value,  100  —  *, 

100*  -  x2  =  2304,     or     a;2  —  100*  =  -  2304. 
whence,  by  the  rule, 


x  =  50  ±y  196  =  50  ±  14; 
hence,  a;  =  64,     x  =  36, 

and  y  =  36,     y  =  64. 

11.  It  is  required  to  divide  the  number  24  into  two  such  parts, 
that  their  product  may  be  equal  to  35  times  their  difference. 

Let     x  and  y     denote  the  parts. 
From  the  conditions  of  the  problem, 

x  +  y  =  24     •     •     .     (1) 
*y  =  35(*-y)    .     .     .    (2); 
substituting  in   (2)  the  value,     y  =  24  —  *, 

24*  -  x2  =  35  (2a:  -  24),     or,     24*  -  *2  =  70*  -  840 ; 


182.1  EQUATIONS  OF  THE  SECOND  DEGREE.  93 

whence,       v  &  +  46a;  =  840  j 

by  the  rule, 

x  sa  —  23  ±  ^840  +  529  =  —  23  =fc  37  ; 

hence,  taking  the  upper  sign,     x  =  14 ; 
by  substitution,     in  (1),  y  s=  10. 

12.  What  two  numbers  are  those,  whose  product  is  255,  and  the 
sum  of  whose  square  is  514? 

Let  x  and  y  denote  the  numbers. 

From  the  conditions,         xy  =a  255     •     •     .      (1) 

x2  f  y*  =  514  .     .     (2), 

multiplying  both  members  of  (1)  by  2,  adding  and  subtracting  the 
resulting  equation  to  and  from  (2),  member  by  member, 

x*  +  2*y  4-  y2  =  1024     .     .     (3) 

x*-2xy+i/  =  4:      .     .     ;     (4); 

extracting  the  square  root  of  both  members, 

x  +  y  =  32, 

*-y  =  2, 

whence,  x  =  17 ;         y  sa  15. 

« 

13.  There  is  a  number  expressed  by  two  digits,  which,  when 
divided  by  the  sum  of  the  digits,  gives  a  quotient  greater  by  2  than1 
the  first  digit ;  but  if  the  digits  be  inverted,  and  the  resulting  num- 
ber be  divided  by  a  number  greater  by  1  than  the  sum  of  the 
digits,  the  quotient  will  exceed  the  former  quotient  by  2 :  what  is 
the  number  1 


94  KEY  TO  DAVIES'  BOURDON.  [182. 

Let  x  and  y  denote  the  digits ;  then  will 

10#  +  y  denote  the  number. 
From  the  conditions. 


x  +  y 

(i) 

(2); 

clearing  of  fractions,  and  reducing, 

8x  —  y    —  x2  —  xy  =  0     • 

•    •    (3), 

6y  —  4x  —  x2  -—  xy  s=  4     *     • 

•   W; 

by  subtraction, 

7y  —  12z  =  4  ;        .  \     y  =  - 

12x  +  4 

7       ' 

substituting  in  (3), 

8.-18s  +  4      *'      ,^  +  4a;  =  0) 

clearing  of  fractions  and  reducing, 

x2  —  —  a?  s=  —  — -,     whence, 

I 

20  /       4    ,   400       20       18 

19      V       19^361       19       19 

Taking  the  upper  sign,  gives     x  =  2 ;     whence,     y  —  4. 

14.  A  regiment,  in  garrison,  consisting  of  a  certain  number  of 
companies,  receives  orders  to  send  216  men  on  duty,  each  company 
to  furnish  an  equal  number.  Before  the  order  was  executed,  three 
of  the  companies  were  sent  on  another  service,  and  it  was  then 
found  that  each  company  that  remained  would  have  to  send  12  men 
additional,  in  order  to  make  up  the  complement,  216.     How  many 


183.]  EQUATIONS  OF  THE  SECOND   DEGREE.  95 

companies  were  in  the  regiment,  and  what  number  of  men  did  each 
of  the  remaining  companies  send  % 

Let  x  denote  the  number  of  companies,  and 

y       "        "         "  each  should  send ;  then, 

y  j^.  12     will  denote  the  number  sent  by  each. 

From  the  conditions  of  the  problem, 

sy  =  216     •     •     •     (1), 

(x  -3)   (y  +  12)  =  216     .     .     ."    (2)  ; 

Performing  operations,  subtracting  and  reducing, 

4z-y  =  12.  .'.       y  =  4x-12; 

substituting  in  (1),      4x2  —  \2x  =  216,     or    x2  —  Sx  =  54 ; 


whence,  *  =  "2  =fc\/+T==3'a'5 

taking  the  upper  sign, 

x  =  9  ;         hence,         y  =  24,     and     y  +  12  =  36. 

15.  Find  three  numbers  such,  that  their  sum  shall  be  14,  th© 
sum  of  their  squares  equal  to  84,  and  the  product  of  the  first  and 
third  equal  to  the  square  of  the  second. 

Let  rr,  y  and  z  denote  the  numbers. 

From  the  conditions  of  the  problem, 

x   +y   +  *    ==14     .     .     •     (1), 
x2  -f  y2  +  z2  =  84     •     •     •     (2), 
xz  =  y2     .     .     .     (3). 
Multiplying  both  members  of  (3),  by  2,  adding  to  (2)  and  reducing 


\ 


96  KEl  TO  DA  VIES*  BOURDON.  [183. 

z2 -f- 2zz  +  z2  =  84  +  y2 ;         .-.     x  +  «  =  -y/84  +  y2  '  '  (4) 
from  (1),  ar-f  0  =  14  — y  •  •  •  •  (5); 

equating  the  second  members  of  (4)  and  (5)  and  squaring, 

84+y2=  196-28y +  y2;         .'•     V  =  4. 
Substituting  in  (1)  and  (3), 

x  f  z  =  10     .     .     .     .     (6) 
#z  =  16     .  *.     x  ss  — . 
Substituting  in  (6)  and  reducing, 

s2  -  10«  =  -  16 

z  =  5  =fc  t^bcs  5  rfc  3, 

2  =  8;         8  =  2, 

and  by  substitution,        ar  =  2  ;         z  =  8. 

16.  It  is  required  to  find  a  number,  expressed  by  three  digits, 
such,  that  the  sum  of  the  squares  of  the  digits  shall  be  104 ;  the 
square  of  the  middle  digit  to  exceed  twice  the  product  of  the  other 
two  by  4  ;  and  if  594  be  subtracted  from  the  number,  the  remainder 
will  be  expressed  by  the  same  figures,  but  with  the  extreme  digits 
reversed. 

Let  x,  y  and  z  denote  the  digits ; 

then,  lOOx  -f-  lOy ■-+■  x    will  denote  the  number. 

From  the  conditions  of  the  problem, 

x*  +  y2  +  z2  =  104     .     .     .     (1) 

y2  -  2xz  =  4         .     .     .     (2) 

100*  +  10y  +  z  —  594  =100*  +  10y  4-  x    (3) ; 


183.]  EQUATIONS  OF  THE  SECOND  DEGREE.  97 

subtracting  (2)  from  (1),  member  from  member, 

x2  -f  2z*  +  z2  sa  100         .  • .         x  +  z  =  10; 
reducing  (3)  x  —  z  =  6  ; 

hence,  a;  =  8,     and     2=2. 

By  substitution,        y  =  6,     and  the  number  is  862. 

17.  A   person  has  three  kinds   of   goods   which   together    cost 

$230-55|-.     A  pound  of  each  article  costs  as  many  Jj  dollars  as  there 

are  pounds  in  that  article  :  he  has  one-third  more  of  the  second  than 

of  the  first,  and  3J  times  as  much  of  the  third  as  of  the  second . 

•  How  many  pounds  has  he  of  each  article  ? 

Let  #,  y  and  z  denote  the  number  of  pounds  of  each  article. 

From  the  conditions  of  the  problem, 


x2        y2       z2   _55255 
24  +  24  +  24  ~"    24   ' 

or, 

x2  +  y2  +  *2  =  5525     (1) 

4 

y2=™x2     .     .     .     (2) 

7                          14 
*  =  2y       •*•     "*?* 

and, 

f-flU    •   •   •   (3); 

substituting  in  (1)  and  reducing, 

x2  =  225  .  • .  x  =  15, 

substituting  in  (2)  and  (3), 

y  =  20  ;  z  =  70. 

18.  Two  merchants  each  sold  the  same  kind  of  stuff:  the  second 
sold  3  yards  more  of  it  than  the  first,  and  together,  they  received 
35  dollars.  The  first  said  to  the  second,  "  I  would  have  received  24 
dollars  for   your  stuff."     The  other  replied,  "And  I  would  have 

7 


98  KEY  TO  DAVIES'  BOURDON.  [183. 

received   12 J  dollars  for  yours."     How  many  yards  did  each  of 
them  sell  ? 

Let  s  and  y  denote  the  number  of  yards  sold  by  each. 

24 

Then  will  —  denote  the  price  the  first  received  per  yard, 

y  \ 

25 

and  —  will  denote  the  price  the  second  received  per  yard. 

From  the  conditions  of  the  problem, 

—  +^  =  35,     or,     48*2  +  25y2  =  70xy  : 

substituting  in  the  second  equation  the  value  of  y  taken  from  the 

first, 

48s2  4-  25  (s2  +  6x  +  9)  as  70s2  +  210s ; 

reducing,  x2  —  20s  =  —  75  ; 

whence,  s   as  10    =fc  <\/25  =  10  ±  5, 

or,  s   =  15 ;  s  as  5 ; 

substituting,  _  y  =  18  ;  y  =  8.     J 

19.  A  widow  possessed  13000  dollars,  which  she  divided  into 
two  parts,  and  placed  them  at  interest,  in  such  a  manner,  that  the 
incomes  from  them  were  equal.  If  she  had  put  out  the  first  portion 
at  the  same  rate  as  the  second,  she  would  have  drawn  for  this  part 
360  dollars  interest ;  and  if  she  had  ^placed  the  second  out  at  the 
same  rate  as  the  first,  she  would  have  drawn  for  it  490  dollars 
interest.     What  were  the  two  rates  of  interest  % 

Let  s  and  y  denote  the  rates  per  cent. 

-Let  z  denote  the  1st  portion  ;  then  will  13000  —  z  denote  the  2d. 


183-113-17.]  EXTRACTION  OP  ROOTS.  99 

From  the  conditions  of  the  problem, 

xz        (13000  —  z)y  ?-  /1X 

m  =  - — Too- ^        or'        **  =  130<%-^    •    •    •    (1), 

j^  =  360,  or,        ^  =  36000  ■     fy.    (2), 

<13(K^|-*)«  -.  490,         or,         13000*  -  zx  =  49000  .     .     (3). 

Substituting  in  (1)  the  values  of  zy  and  zx  taken  from  (2)  and  (3) 
and  reducing,  we  find,  x  =s=  y  +  1. 

Substituting  this  value  of  #  and  the  value  of  z  taken  from  (2)   in 
(1),  and  reducing,  we  find  '      , 


,       72  36 


36  /36       1296       36  ±  42 


36         im      U 
Whence,       y   ^^y^-^ 

by  substitution, 


13       '        *    *     V  ~ 


#k7,     and     z  =  6000,     13000  -  z  =  7000* 

EXTRACTION  OF  ROOTS. 
2.  Ans.     x2  +  2x  —  4.         3.  -4n.?>     2z2  —  x  -f  2. 

5.  <4nA     3a^  —  2ta.  6.  -4ns»      2^  —  1,  7 

TRANSFORMATION  OF  RADICALS. 

\ 

ADDITION. 

1.  V4SP  =  45v/3a,     and   b^Tba^hby/Za -y    .%%   ilwj.  9^0, 

2.  3^4^=    3^/2a,       ;       2^2^  .*.       Ans.    5^2^ 

3.  2V45sn    6^        ;       3V57  .\      ^4rcs.    9^5. 


OO'  Of  TH1 


100 


KEY  TO  DAVIES*  BOTTRfcON, 
SUBTRACTION. 


[218-19-20. 


1.  \fia*b  +  16a*  =s  2a »/5  -f-  2a,     and 

^6«-f  JW    -    ^V^  +  2a,      .-,     .4n*.  (2a  -  5)  %/b+*i> 
3.  3^/ia^  =  3^/2^  ;    23y/2a~;      ,'.       Jnt.  ySi, 

MtfLTHTLlCATlOff, 

5.  </2  X  y3  =  l2y64x  12^/81  =  iy5184 

Vr   3  /r  i2  /F  i2  /f  i2  / 1  >-; 

§xvr  vs*  Vsr^  vW  :;:;:f*#? 

6.  2-v/l5 -26^/3375;  3^10^3^/100;  .-.  Jn*,  6^337500 
8.       V^V5*;     3y/3^Y^5     V^-V1^ 


^w*.  jy648000. 


9'  Vr    V  ?29  '    V2        V  16384'     Vb-   V~lb> 

42      /g 

.  * .     Ans.    *  /  — 


10.  The  product  by  the  rule  for  multiplication  is 
28  ft      „     J%      10      43   .    in     Ft      43  ,   13     ~ 


DIVISION. 


2.     2^3  x  ^/i"  -  2  ^y^y  X  ^256  -  2  »/720  x  256, 

12    /729  v  256  

2i2v/7^9x256-f-i12/8x81=£::4    V       8  x  81   ^lV^  An9> 


220-21.] 


TRANSFORMATION   OF   RADICALS. 


101 


3.  y/V*x2>/3  =  yV«  x2V9=\/26V/«  =  12\/^ 


yfi\/2  X^/3  =  2^/^/4  X  V27  =  2  V10^ 
/ a 1  12  Pv*T      1 12  /2 

-  - /9  —    Ans. 


Ans. 


V-5 


2  = 


10 


5.     — = =:  multiplying  both  terms  by  i/a  —  */6,  we  have 

Vs  +  V6  1 

l/a  — -y/6 
again  multiplying  both  terms  by   -^/a  -f  ^/b,  we  finally  obtain 


V°3  ~  VaH  +  Va62  ~~  V63 


Ans. 


a  —  b 

multiplying  both  terms  by  */a  -f  t/^  we  have 

\/a  +  2  4v/o6  +  -/^     and     <s/a—y/b; 
multiplying  both  by   -y/a  -f  -y^  and  we  have 

a  +  ^  +  2y^64-2V^-f  2t/a^ 


a  —  b 


Ans* 


i       r>  i  2-v/2"x  (3)3  .      , 

1.     Keduee    ■  to  its  simplest  terms. 


iV* 


102  KEY  TO  DA  VIES*  BOFEDON.  [232-233. 

Cancelling  the  ^2,  and  writing  the  quotient  of  2  -f-  J,    we  havt* 

{  4(2)^3/3  )  * 
- ¥—-  y      to  its  simplest  terms. 

Raising  both  terms  to  the  4th  power,  we  have 


(1)«3»3»  __  ft)*  (3)  V3_       ys  1 


(2)*2(3)a       (2)3(3)2      •      (2)7x3       384 


/(  a  Y3  _L  .  /ai  )  "s 


(JV3  _L    -/gj. 

3.     Reduce     /  J  — ~-~|  V      to  its  simplest  terms. 

v  ( »%$$>*  j 

Since  the  square  root  of  the  square  root  is  equal  to  the  4th  root, 
we  need  only  operate  on  the  terms  of  the  fraction  : 

,    (F  +  yi?^  i_+yy'=*+yT 

2-/2x(3)*      *^*P§  ^6 

Multiplying  both  terms  of  the  fraction  by  the  -\ft$,  we  have 

I  ih£±^£3$3d+m  :  hence, 


4.  Multiply     a*+  a24*  +  aV  +  aJ  +  o¥  +  6  , 


."2t?-  i     «i#  ,  Jb  ,     it 


1     5 


a3  -f  aJ&rf  +  a2b*  +  azb  -f  a&*  -f  a"^ 

_  A*  -  a?$  -  ah  -  ab*  -  gM  -  jg 
a3  —  62.     ^4t?s. 


l?-241-42.]             ARITHMETICAL    PROGRESSION. 

$ 

5.     DLvide    a*-a*A"*-A+&*     ||a*-6'* 

a*  _  a?b% 

&-b 

-ah  +  b* 
-  ah  +  6*. 

103 


ARITHMETICAL  PROGRESSION. 

2.  a  =  2,     rf  =  7,     ?i  =  100;"    .  \   I  =  a  +  (n  -  l)d  =695. 

3.  a  =  1,     a7  =  2,     n  =  100  ;     . ' .    J  =  a  +  (n  —  1)  a*  =  199. 
Hence,  S  =  %n  (a  +  I)  =  10000. 

4.  Z  =  70,     c?  =  3,     n  =  21, 

a  =  I  -  (n  -  1)  d  =  10  ;     S  =  }jn  [21  -i  (n  -  1)<]  =840'. 

5.  a  =10,     d=-$,     n  =  21     .  •.  /  =  a  +  (w  -  l)rf  =  *£■', 
whence,  S  =  }n  [21  -  (n  -  l)rf]  =  140. 

6.  a*  =  6,     J  s  185,     S  =  2945 


_  2Z  +  d  db  ygg  +  rf)2  -~8rf£  _  376  ±4 

taking  the  lower  sign,     n  =  31,         a  =  /  —  (rc  —  l)a*  =  5. 

7.  a  =  2,     Z  =  5,-    n  =  llj     .  • .  d  =  ^*  =  ~  =  0.3. 

n—  1       10 

8.  a  =  1,     a7  =  1,     n  =  7i      7  =  n  ; 

...     ^=in[2a+0i-l)d]  =  (i-±^. 

9.  a=l,     d  =  2,     »==n;     .-.   5  =  \n  [2a  +  (»  —  l)d]  =  n» 
10.     a  =  4,     c?  =  4,     n  =  100  ; 

.   .     8  =s$n  [2a  +  (n  -  l)d]  =  20200 
20200yck.  =  llmt.  840ya7s. 


104  key  to  davies'  bouedon.  [245-46-48-67-68. 

GEOMETRICAL   PROGRESSION. 

3.  a  =  2,     r  =  3,     /  =  39366, 

8  as  ^-^  =  59048. 
r  —  1 

4.  assl,     r  =  2,     n  =  12; 

(IT*   — —  ffl 

.  •.     £  = =  4095  ;     I  =  ar»-i  as  2048. 

r  —  1 

5.  a=l,     r-2,     w  =  12; 

.  • .     S=  ar*  ~?  =  4095  ;  4  195*.  =  £204  15*. 
r  —  1 

6.  a  =  1,     r  =  3,     n  =  10  ; 

_       ar»  —  a       59048         .^  ^ 

.-.     S  = =—^—    295.24. 

r  —  1  2 

/  =  ar"-1  =    196.83. 

INSERTING    GEOMETRICAL    MEANS. 

2.     az=2,     &  =  486,     w  =  4;         .-.     r  =  ^243  =  3: 
hence,  the  progression,         2  :  6  :  18  :  54  :  162  :  486. 

SUMMATION   OF  SERIES. 
4.  p  =  4,     q  =  4,     n  =  1,     5,     9,     13,     &c. 

4       4      4        4        4 

1st  auxiliary  series,     --f^-f---f  —  +  _+... 

4       4        4         4  4 


5   '   9       13   '    17    '  r  4» -3' 


^  =  i(4-i^l)-'if    "=»■     *** 


274-286.]  EXPONENTIAL    EQUATIONS.  105 

PILING  BALLS. 

l  _15;    ...S  =  l^+Jl^±3  =  660- 

i     n  =  U.       ...ig="<"+|1>^.±l>=10i6; 

1 .  Z  .  o 

f     ,  »'  =  5,     S'  =  55  ;       .  • .   S  —  S'  =  960. 

3.  m  =  30,    -  =  30  ;      .  ,  8  =  ^^f+H,  23405. 

4.  m  =  26,     n  =  20  ;        .  • .   S  =  8330  ;     m  =  26,     ft'  =  8  ; 

S'  =  1140;      .-.  fif-  £'^7190. 

5.  »  =  20;     .-.5=1540;     ft'  =  9 ;     .-.  S'  =  65; 

...  £-£'  =  1475. 

6.  w  =  15;     .-.5=1240;     »' =  5 ;     .-.5' =  55; 

.-.   S-  8'  =  1185. 

7.  w=  52,     ft  =  40  ;     .  • .  S  =  G4780  ;     m  =  52,     n'  =  18 ; 

.  • .  fif  =  11001  ;     .-.£-£'  =  53679. 

EXPONENTIAL  EQUATIONS. 
These  equations  may  be  solved  as  the  preceding  ones  have  been, 
but  it  will  be  better  to  make  use  of  the  table  of  logarithms  (P.  288), 
together  with  the  principle  that  the  logarithm  of  a  power  is  equal  to 
the  exponent  of  the  power  into  the  logarithm  of  the  root. 
This  principle  gives  the  following  solutions  : 
I-  3*  =  15;  %  )(«.     fQ 

taking  the  logarithms  of  both  members, 

i      o       i      k  log  *       1.176091       _    Ad 

slog3  =  log  15;       .v*  =  ^  =  5-^—  =  2.46... 


106  KEY  TO  DAVIES'  BOUKDON.     '  [286-309-310. 

2  10*  =  3; 

taking  the  logarithms  of  both  members, 

x  log  10  =  log  3,     or    x  =  log  3  =  0 .  477121- 

3.  5*  =  J; 

taking  the  logarithms  of  both  members, 

x  log  5  =  log  f  =  log  2  —  log  3  ; 

..  log2-log3_  -.176091    _ 

log  5        ~       .698970  ~  "~  "^ 


THEORY  OF  EQUATIONS. 

2.  Two  roots  of  the  equation, 

x*  —  12a;3  +  48a:2  —  68a;  +  15  =  0, 

are  3  and  5  :  what  does  the  equation  become  when  freed  of  them  ? 

**  -  12a;3  +  48a>2  -  68a-  +  15  Ifj-j 

X2  _  [)X2  4-  21a;  —  5 

x2  -    9a;2  4-  21a;   —  5  |  x   -  5 

x2  —  4a;  +  1  =0.     Ans. 

3.  A  root  of  the  equation, 

X3  _  6a-2  -f  11a; -6  =  0, 

is  1  :  what  is  the  reduced  equation  1 

xs  —  Qx2  +  11a;  —  6  |  x  —  1 

x2  —  bx  4-  6  =0.    Ans. 

4.  Two  roots  of  the  equation, 

4a;4  —  14a;3  —  5a;2  4-  31s  +  6  =  0, 
are  2  and  3  :  find  the  reduced  equation. 


314-325.]  GREATEST   COMMON   DIVISOR.  107 

4**  -  14*3  —  5*2  +  31*  +  6  |  *  -  2 

4*3  —  6*2  -  17*  —  3, 

4*3  —  6*2  —  17*  -  3  j*-3 

4*2  +  6*  4-  1  =  0.     -dws 

FORMATION  OF  EQUATIONS. 

2.  What  is  the  equation  whose  roots  are  1,  2,  and.  —  3  ? 

(*  -  1)  (x  -  2)  (x  -f  3)  =  *3  -  Ix  +  6  =t  0.     Ans. 

3.  What  is  the  equation  whose  roots  are  3,    —  4,    2  -f-  %/**>  an(^ 
2-  ^fl 

(*  -  3)  (x  +  4)  (*  -  2  -  V3)  (*  -  2  +  -v/3) 

'  =  **  —  3*3  —  15*2  +  49*  —  12  =  0.     Ans. 

4.  What  is  the  equation  whose  roots  are    3  -+•  \/5^  3  —  ^/b[ 
and  —6? 

(x-3  -  Vo)  (x  -  3  +  <v/5)  (*  -f  6)  =  *3  -  32*  +  24  =  0.    Ans. 

5.  What  is  the  equation  whose  roots  are    1,-2,  3,  —  4,  5, 
and  —  6 1 

(x  -  1)  (* +  2)  (x  -  3)  (*  +  4)  (x  -5)  (x  +  6) 
=  *6  +  3*5  —  41**  —  87*3  +  400*2  +  444*  —  720  =  0.     Ans. 


6.  What  is  the  equation  whose   roots   are  ....  2  +  y—  1, 


2—  y  — 1,  and   -3? 

(z  _  2  - yCTT)   (*  -  2  +V=r\)  (*  +  3) 
=  *3  —  *2  —  7*  +  15  =  0.     Ans. 

GREATEST  COMMON  DIVISOR. 
1.  Find  the  greatest  common  divisor  of  the  two  polynomials, 
r  6*5  —  4*4  -  1 1*3  —  3*2  —  3*  —  1, 

4**-f  2*3--  18*24-3*  —5. 


108  KEY  TO  DAVIKS'  BOURDON.  [325. 

Multiply  the  first  polynomial  by  2  to  make  the  first  term  divisible 
by  the  first  term  of  the  second  ;  thus, 

12*5  -  8a;4  -  22a;3  -  6x2  -  6x  -  4  |  4a;4  +  2a3  -  18a;2  +  Sx  —  5 

3a:,  7,    a  remainder. 

78a;3  —  156a;2  -}-  39a;  —  39,  and  suppressing  common  factor    39, 

2*3  _      4^2  +      x  _    i  .     theil5 

4a;4  +      2a:3  —  18a:2  -f  Sz  -  5  |  2a:3  —  4a:2  +  x  —  1, 

~2a^       51 

hence,  2a:3  —  4a:2  -f-  #  —  1,      is  the  greatest  common  divisor. 

2.  Find  the  greatest  common  divisor  of  the  polynomials, 

20a:6  -  12a:5  -f  16z4  -  15a:3  +  14a:2  —  15a;  -f  4,       ' 

and       15a;4  —    9a;3  +  47a:2  —  21a;   +  28. 

Multiplying  the  first  polynomial  by  3,.  and  performing  the  division, 

60a;6  —  36z5  -f  48a;4  —  45a;3  +  42a:2  —  45z 

-f  12  |  15a;4-9a;3-f47a;2-21a;  +  28 


4a;2,  28 

60a;6  -  36a-5  4-  188a:4  -  84a-3  +  112a;2 

-  HOz4  +  39a:3  —  70a;2  —  45a;  +  12. 

Multiplying  the  remainder  by  3,  to  make  the  first  term  divisible, 

—  420a;4  +  117a;3  -210a:2    -  135a;  4-  36 

-  420a;4  -f  252a;3  -  1316a:2  +  588a;  -  784 

—  135a;3  -f  1106a;2  —  723a;  +  820.     1st  rem. 

Multiplying  the  last  divisor  by  9,  and  then  using  it  as  a  dividend, 

135a;4-81a;3+423a;2-lS9a:+252  |  -135a:3+ 1106a:2— 723a; +820 

-a-,  -205 

135a:4  —  1106a:3  +  723a2  —  820a; 

~   -u  1025a:3  -  300a;2  +  631a;  +  2527 


326-335,]  GREATEST  COMMON   DtVlSOK,  109 

multiplying  the  remainder  by  27,  we  have 

27675a;3  -      8100a:2  +     17037a:  +     6804 

27675*3  -  226730a;2  +  148215a  —  168100 
218630a:2  —  131178a:  +  174904} 

dividing  by  the  common  factor  43726,  we  have 

5^2  _  ^x  ±  4, 

which  will  divide  the  last  divisor,  and  is  the  greatest  common  divisof 
«ought« 

3*  Find  the  greatest  common  divisor  of  the  two  polynomials 

5a462*  +  2a363  +  cat   ~  3a264  +  bca, 

and  afi  -f  5add  —  a*b2  +  ha?bdt 

Since  a  is  a  factor  of  both  polynomials,  we  set  it  aside  as  forming  a 
part  of  the  common  divisor  :  then  we  have 

5a362  -f  2a2b3  -j-  ca  ~  3ab*  -f  be, 

and  a*  -f  ba\l  ~  aW  -f  5aW  ; 

observing  that  c  does  not  enter  the  second  polynomial,  we  arrange 
the  first  with  reference  to  it,  and  have 

(a  +  b)c  +  (5a362  -f  2a*b3  —  Sab4)  c°  : 

the  common  divisor  of  the  two  last  polynomials  is  a  -f-  b  (Art  261, 
Bourdon)  :  hence,  tf3  +  <&  »  the  common  divisor  sought. 

TRANSFORMATION    OF   EQUATIONS. 

4.  What  is  the  transformed  equation,  if  the  second  term  be  made 
to  disappear  from  the  equation 

xb  _  io  z*  4  7a:3  4*4r-9s0l 


110.  KEY  TO  DAVIES*  BOtTRDON.  [335. 

Make   (Art.  263,^  «  =  «  +  -  =  «  + 2;         nence,     z'  =  2, 

D 

and  the  transformed  equation  will  be  of  the  form 

X  =         (2)6~     10  (2)* -f    7(2)3  +  4(2)»-9,  or,  1^-    73 
F  as      5  (2)*  -     40  (2)3  -f  21  (2)2  +  4  F  =  -  152 

2}  =    20  (2)3  -  120  (2)2  +  42  (2)  ~  =  -  118 

P=    60  (2)2-  240(2) l +  42 ^L  ~  _    33 

1  • « .  o 

r=  120  (2)1  -  240 x £      =       0 

f/=120    • •    •    •       •     172^7475=         » 

us  _  33tt8  _  ii8»3  _  i52lt  -.  73  =  0. 

6.  What  is  the  transformed  equation,  if  the  second  term  be  made 
to  disappear  from  the  equation 

3z3  +  15a.-2  -f-  25z-3  =  0? 
Reducing  the  co»efficient  of  the  highest  power  of  the  unknown 

quantity  to  1, 

2t 

z3-f  5z2  +  —  x-l  =0. 

o 

5  5 

Make,  p  a  u  —  - ;     wherjce>     ar  sb  —  - ; 

and  applying  the  above  form,  we  have 

3       15'2       A         A 
tt3  -—  =  0.      An*> 


335-41.]  EQUAL  ROOTS.  Ill 

6.  Transform  the  equation 

3a;4  -  13a;3  +  7x2  —  8a;  -  9  ^  0 
into  another,  the  roots  of  which  shall  be  less  than  the  roots  of  the 

given  equation  by  -• 
Make,    x  =  u  -f-  - ;  whence,     x'  =  +  - ; 

o  o  t 

applying  the  above  formula,  we  have, 

\ 


EQUAL  ROOTS. 

4.  What  are  the  equal  factors  of  the  equation 

xi  _  7*6  +  10*5  +  22a;4  -  43a;3  -  35a;2  -f-  48a;  -f  36  t=z  0 1 

The  first  derived  polynomial  is 

7a;6  —  42a;6  +  50a;4  -f  88a;3  -  129a;2  -  70a;  -f-  48, 

and  the  common  divisor  between  it  and   the  first  member  of  the 
given  equation,  is 

x*  —  3a;3  —  3a;2  -f  7a;  +  6. 

The  equation 

a;4  —  3a;3  -f  3a;2  +  7a;  -f  6  —  0, 

cannot  be  solved  directly,  but  by  applying  to  it  the  method  of  equal 

roots;  that  is,  by  seeking  for  a  common   divisor  between  its  first 

member  and  its  derived  polynomial, 

4a;3  —  9a;2  —  6x  +  7, 

we  find  such  divisor  to  be  x  -f  1  ;  hence,  x  +  1  is  twice  a  factor  of 

the  first  derived   polynomial,  and  three  times  a  factor  of  the  first 

member  of  the  given  equation  (Art.  268). 


112  KEY  TO  DAVIES'  BOURDON.  [341. 

Dividing, 

x*  -  3^3  _  3^2  _f_  lx  +  6  _.  o,     by     (x  -f-  1)2  e&  x*  +  2x  +  1, 

We  have,  a;2  —  5a;  -p*  6, 

which  being  placed  equal  to  0,  gives  the  two  roots 

xs2         and         x  x=  3, 

und  the  two  factors,    x  —  2         and         a:  —  3. 

Therefore,  (a;  —  2)  and  (re  —  3),  each  enters  twice  as  a  factor  of  the 
given  equation ;  hence,  the  factors  are 

(x  -  2)2  (x  -  3)2  (x  +  1)3.     Am. 

6.  What  are  the  equal  factors  of  the  equation 

x*  ~  <sxe  +  Dzfi  _  19^4  +  27a:3  -  83*»  -f  27a:  -  9  =  0? 

The  first  derived  polynomial  is 

7a:6  -  18a,-6  +  45z4  -  76a:3  -f  81a2  -  66a;  +  27, 

and  the  common  divisor  between   it  and  the  first  member  of  the 
given  equation,  is 

a.4  _  2a:3  +  4a:2  -  (jx  -f  3. 

The  equation 

^4  _  2a:3  +  4a:2  -  6a;  -f  3, 

cannot  be  solved  directly,  but  by  applying  to  it  the  method  of  equal 
roots,  as  in  the  last  example,  we  find  the  derived  polynomial  to  be 

4a;3  —  6a:2  +  8a;  —  6 
and  the  common  divisor  to  be  x  —  1 ;  hence,  (x  —  1)  enters  twice 
as  a  factor  into  the  derived  polynomial,  and  three  times  as  a  factor 
into  the  first  member  of  the  given  equation. 
Dividing 

x*  -  2x  +-  4z2  ~  6.r  +  3     by     (x  -  l)2  sr  a;2  -  2a;  +  1, 


358-370.]  SMALLEST  LIMIT  OF  ENTIRE  ROOTS.  118 

we  have  for  a  quotient  x2  +  3  ; 

hence,  (s2  +  3)  enters  twice  as  a  factor  into  the  first  member  of 

the  given  equation  :  hence  the  factors  are 

(x  -  l)3  (s  +  3)2. 

SMALLEST  LIMIT   OF  ENTIRE  ROOTS. 

2.  What  is  the  superior  limit  of  the  positive  roots  in  the  equation 

X5  __  3xa  __  $x3  __  25s2  -f  4s  -  39  =  0 1 
Recollecting  that  if  we  use  s  for  x'  (Art.  285),  we  have 
X  =    x5  —  3s*    —  Sx3   —  25a;2  -f  4s  —  39 
Y  =  5s*  —  12s3  -  24s2  -  50s  +  4, 

Z  =  20s3-  36s2  -  48s  -  50, 

V=  60s2-  72s  -48, 
W=  120s  -  72. 

T=z  120 

The  least  whole  number  that  will  render  all  these  derived  polyno- 
mials positive  is  6 ;  hence,  6  is  the  superior  limit. 

3.  What  is  the  superior  limit  of  the  positive  roots  in  the  equation 

s5  -  5s*  -  13s3  -  17s2  -  69  =  0. 

The  process  of  solution  is  the  same  as  in  the  last  example,  and  the 
limit  is  found  to  be  7. 

COMMENSURABLE    ROOTS. 

2.  What  are  the  entire  roots  of  the  equation     - 
s*  —  5s3  +  25s  —  21  =0?' 
The  divisors  of  the  last  term  are  -f  1,  —  1,  +3,   —  3,  -f  7,  —  7^ 
-f  21,  and   -21  :  X  =  22;   -  Z"=  -4. 
8 


114  KEY  TO  DAVIES'  BOURDON.  [370. 

+  21,     +7,     +3,     +1,  -1,  -3,     - 

-    1,     -3,     -7,     -21,  +21,  +7,     + 

+  24,     +  22,  +  18,  +    4,  +  46,  +  32,  + 

+    3,  +    6,  +    4,  -  46,  - 

+    2,  +    4,  +46, 

-3,-1,  +  41, 

-   i,  -   i;   -41, 

therefore,    +  3  ancl   +  1   are  the  two  entire  roots.     Dividing  the 

first  member  of  the  equation  by  the  product  of  the  factors 

(x  -  3)   (x  -  1)  =  z2  -  4x  +  3, 

we  have  x2  —  x  —  y  =  0. 

Note. — In  the  4th  line  we  add  the  co-efficient  of  #*,  which  is  0,  and  then 
divide  by  the  divisors,  and  thus  obtain  the  6th  line. 

3.  What  are  the  entire  roots  of  the  equation 

15z5  -  ID*4  +  6z3  +  15z2  -  19#  +  6  =  0  ? 


+    6, 

+ 

3, 

+    2, 

+       1, 

-    1, 

-    2, 

-    3, 

-    6, 

+    1, 

+ 

2, 

+    3, 

+    6, 

-    6, 

-    3, 

-    2, 

-    1, 

-18, 

— 

17, 

-16, 

-13. 

-25, 

-22, 

-21, 

-20, 

-    3, 

-    8, 

-13, 

t-25, 

+  11, 

+    7> 

+  12, 

+    1, 

+    2, 

+  40, 

+  26, 

+  22, 

+    2, 

+    2, 

-40, 

-13, 

+    8, 

• 

« 

+    8, 
+    8, 
-11, 
-11, 

-34, 
+  34, 
+  15, 
-15, 
+  15: 

-    7, 

hence,  there 

is  but  one  entire  root 

,  which  is  —  1. 

370-378.]  NUMBER  OF   REAL  ROOTS.  115 

4.  What  are  the  entire  roots  of  the  equation 

9z6  -f  30a:5  -f  22tf  +  10a;3  +  17a:2  —  20a;  -f  4  =  0  ! 
This  example  is  worked  like  the  preceding,  and  there  is  but  two 
entire  roots. 

KUMBER  OF  REAL  ROOTS. 

3.  What  is  the  number  of  real  roots  of  the  equation 

X3  _  5*2  +  Sx  _  i  =  o  | 

By  finding  the  expressions  which  indicate,  by  their  change  of  sign, 
the  existence  of  real  roots  (Art.  302  and  Example  1),  we  have 
X  ss  a3    -  5a:2  -f  8*  -  1, 
Xx  =z  3a;2  —  10a:  +  8, 
X2  =  2x   -  31 
X3  4  -  2295, 

x  'H  —  oo   gives 1 2  variations, 

x  =  +  oo   gives     -f  -f-  H 1  variation  ; 

hence,  there  is  one  real  and  two  imaginary  roots  (Art.  300). 

For      x  as  0,     we  have f- 2  variations, 

for       x  =  1,  "  +  H 1  variation: 

hence,  the  real  root  lies  between  0  and  -f-  1. 

4.  What  is  the  number  of  the  real  roots  of  the  equation 

x*  —  x3  —  2a:2  —  x  —  3  =  0  ? 
X  =  a:4       —  z3  —  2a:2  —  a;  —  3, 
X,  =  4a:3     -  3z2  -  4x  —  1, 
X2  =  19a:2   +  16a:  +  49, 
X3  b£  101a;  -  174, 
X<  =  -  1356277. 


116  KEY  TO  DAVIES*  BOURDON.  [378-381. 

For    x  =  —  oc  ,     we  have         -\ \- ,  3  variations, 

for      »=       oo ,        "  4*  -f  +  H ,  1  variation ; 

hence,  there  are  two  real  roots. 

5.  What  is  the  number  of  the  real  roots  of  the  equation 
xs  __  2z3  +  1  =  0  ? 

X  =  x5  -  2a;3  +  1, 

Xx  =  5**  -  6*2, 

X2  =  4a:3  —  5, 

X3  ae  24a;2  -  25a;, 

X4  =  —  5a;  +  6 

X>  =  -  114. 

For    x  =  —  oo  ,     we  have       - — | f-  -f ,      4  variations, 

for      «=       co ,         "  -f  +  4*  + >       1  variation  % 

hence,  there  are  three  real  roots.       , 

CUBIC  EQUATIONS. 

1.  What  are  the  roots  of  the  equation 

x*  _  6z2  +  3a;  *  18     -     •     \     (1)? 
Transforming  so  as  to  make  the  second  term  disappear, 
a;3_9.T_28=xO     *     •     •     (2) 
p=  —  9         q=z  —  28; 
tubstituting  in  Cardan's  formula,  and  reducing, 

x  =  4. 
But  the  roots  of  the  given  equation  are  greater  than  those  of  equa- 
tion (2)  by  2  ;  hence  x  =  6, 


381.]  CUBIC    EQUATIONS.  117 

Transposing   18   in  equation  (1),  and  dividing   both  numbers  by 
x  —  6,  we  find 

x2  +  3  =  0         .-.         x  =  ±  ^^\ 

hence  the  three  roots  are         6,     y/  —  3,     —  V  ~  3. 

2.  What  are  the  roots  of  the  equation 

^3  _  9;C2  +  28z  =  30     ....     (1)? 
Transforming,  we  find 

x3  +  s  =  0  (2)     .-.     s  =  0    and    a;  as  ±  y^H. 
But  the  roots  of  (1)  are  greater  than  those  of  (2)  by  3; 
hence  the  roots  of  (1)  are,     3,  3  4-  y/  —  1     and    3  —  */--  1. 

3.  What  are  the  roots  of  the  equation 

a3  —  lx2  -f  14*  =  20     •  •     (1)? 

Transforming,  we  have 

•,      7  344      ,         . 


7  344 

*=-3'*=-2^ 


substituting  in  Cardan's  formula,  we  find 

8 
x  =  -  for  the  real  root. 

But  the  roots  of  (1)  are  greater  than  those  of  (2)  by  - ; 

3 

hence,  for  (1)  we  have  x  =  5. 

Transposing  in  (1),  and  dividing  both  members  by  x  —  5, 
•     x2-2x  +  4  =  0; 


118  KEY  TO  DAVIEs'  BOFKDON.  [391. 

or,  x2  —  2x  =  —  4 ; 

whence,  x  =  1  ±  y'—  3; 

hence  the  roots  required  are.    5,  1  +  y  — 3    and     1  —  */  —  3. 

TRANSFORMATION  OF  EQUATIONS  BY  SYNTHETICAL  DIVISION. 

3.  Find  the  equation  whose  roots  shall  be  less  by  1  than  the 
root&of 

x3  -  7*  +  7  =  0. 

1  +  0  -  7  +  7  ||  1_ 

+1+1-6 
+  1  -  6,+  1 

+  1+2 


+  2,-4 
+  1 


l,+  3  .-.  y3_f.33/2_4y  +  1:=0# 

4.  Find  the  equation  whose  roots  shall  be  less  by  3  than  the 
roots  of  the  equation 

a*  __  Sx3  —  lbxz  +  49a:  —  12  =  0. 

1  -3  —  15  +  49  -  12 1|  3 
+  3  +    0-45+12 


+  0-15  + 

4,+    0 

+  3+    9- 

18 

+  3-    6,- 

14 

+  3+18 

+  6,+ 12 

+  3 

l,+  9      .-.       y*  +  9y3  +  12ya  -  14y  =  0. 


391-400.]  horner's  method.  119 

5.  Find  the  equation  whose  roots  shall  be  less  by  10  than  the 
roots  of  the  equation 

'      x*  +  2z3  +  Bx2  +  4x  —  12340  =  0. 

1+2+      3+        4-  123401110 
+  10  +  120  +  1230  +  12340 
+  12  +  123  +  1234,+  0 
+  10  +  220  +  3430 


+  22  +#343,+  4664 


+  10  +  320 
+  32,+  663 
+  10 


l,+  42     .-.     y*  +  42y3  +  663y2  +  4664y  =  0. 

6.  Find  the  equation  whose  roots  shall  be  less  by  2  than  the  roots 

of  the  equation 

x*  +  2x3  —  Qx2  —  lOx  =  0. 

1+0+    2-    6  -  10  +  0  ||  2 
+  2+    4+12+  12'+ 4 


+  2+  6+  6+  2,+ 4 

+  2  +  8  +  28  +  68 

+  4  +  14  +  34,+  70 

+  2  +  12  +  52 

+  6  +  26,+  86 

+  2  +  16 

+  8,+ 42 

+  2 

1,+  10;     .-.  y£  +  10^+422/3+86y2+70y+4=0. 

HORNER'S  METHOD   OF   SOLVING   NUMERICAL  EQUATION& 
1.  x3  +  x2  +  x  —  100  =  0. 

By  Sturm's  Rule,  we  find 


120 


KEY  TO  DA  VIES'  BOURDON. 


[400. 


variations, 
variation ; 

variations, 
variation ; 


X  =  x3+z2  +  x   -  100, 

XI  =       3a;2  -f  2x  +  1 
X2  =  -  4a;   +  899 
X,  =  -  2409336. 

For        z  =  -  go  ,  _  +  +  _}  2 

for  x  =  +  ao ,  -f  4-  —  — ,  i 

hence,  there  is  but  one  real  root. 
Tor        x  =  4,  -  4-  4  -  ,  2 

for  x  =  5,  4-  4-  H ,    -         1 

hence,  the  real  root  lies  between  4  and  5. 

2.  z*  —  12z2  4-  12a;  —  3  =  0. 

By  Sturm's  Rule, 

X  =x*    -  12a;2  4-  12*  -  3, 

Xx  =  4a;3  -  24*  -f  12,     or    z3  -  6a?  4-  3, 

X2  =  2a*  —    3a:  4-  1, 

X3  =  13a;—    9,       • 

X4  =  20. 

a;  =  —  go  ,  H 1 f-  ,  4  variations, 

z  =  4-  co  ,  4-  4-  4.  +  4.  j  q  variation ; 


For 
for 


hence,  there  are  4  real  roots. 


For 

for 

for 

for 

for 

for 


a;=,-4 
xz=  -3 
x=  0 
3=  4-  1 
x=  4-2 
a?  =  4-3 


+  -  4-  -  4- 

+  -  4- 

-  +  +  -  4- 

T  +  + 

4-  +  4- 

+  4-  4-  4-  + 


4  variations, 
3  variations, 
3  variations, 
1  variation, 
1  variation, 
0  variation : 


400.] 


horner's  method. 


121 


.hence,  one  of  the  roots  lies  between  —  4  and  —  3,  two  between  0 
and  1,  and  the  remaining  root  lies  between  2  and  3. 

3.  s*  —  8s3  +  14s2  +  4x  —  8  =  0. 

By  Sturm's  Rule, 

X  =  s4    -  8s3    +  14a2  +  4x  —  8,  n 

Xx  =  4x3  -  24s2  -f  28s  +  4,     or    s?  -  6s2  -f  7s  +  1, 

X2  =  5s2  -  17s  -f-  6, 

X3  =  76s  -  103, 

X4  rr  45475. 


For 

a  =  —  co ,         H 1 f- 

4  variations, 

for 

z=  +  ao,         +  +  -f  +  + 

0  variation ; 

hence,  the 

equation  has  4  real  roots. 

For 

s  =  -l              -r  -  +  -  + 

4  variations. 

for 

s=       0             -  +  +  -  + 

3  variations, 

for 

x=  +  1              +  + + 

2  variations, 

for 

x=+2             +  -  -  +  + 

2  variations, 

for 

IB  +3              ±  +  + 

1  variation, 

for 

x=       5              +  4  + 

1  variation, 

for 

s=       6             +  +  +  +  + 

0  variation ; 

hence,  one 

j  root  is  between  —  1   and  0,  one  between  0  and  1,  one 

between  2  and  3,  and  one  between  5  and  6. 

4. 

s5  —  10s3  -f  6s  +  1  =  0. 

By  Sturm 

,'s  Rule, 

X  =  s5      -  10s3+  Qx  +  1, 
X,  =  5s*    -  30s2+  6, 
X2  =  20s3  -  24s  -  5, 
X3  =  96s2  -  5s    -  24, 
X4  =  43651s  +  10920, 
X5  =  32335636224. 

122 


KEY  TO  DAVIES    BOURDON. 


1400. 


For     x  =  —  go          h h  —  +  5  variations, 

for      #=  +  00          +4-  +  H — h  +  0  variation ; 

hence,  the  equation  has  5  real  roots. 

h h f-  5  variations, 

+  H h  —  -f  4  variations, 

+ h  —  4-  4  variations, 

+  + h  -H  2  variations, 

1 — h  +  1  variation, 

h  4-  H — h  1  variation; 

+  +  +  +  H — h  0  variation  ; 

hence,             one  root  lies  between     —  4     and  —  3, 

two  roots  lie  between     —  1     and  0, 

one  root  lies  between          0     and  +  1, 

one  ro>t  lies  between         3    and  4. 


For 

x=  -4 

for 

x=  -3 

for 

x=  -1 

for 

**=       0 

for 

*  =  +  l 

for 

x=  +3 

for 

x—  +4 

APPENDIX. 


GENERAL  SOLUTU.rf  OF  TWO  SIMULTANEOUS  EQUATIONS  OF  THE 
FIRST  DEGREE. 
1.  Take  the  equ&cions, 

ax  +   by  =  c      .     .     .     (1), 
a'x+b'y  =  c'     ...     (2); 

multiply  both  members  of  (1)  by  V  and  of  (2)  by  6,  then  sub- 
tracting and  factoring,  we  find 

(ab'  —  a'b)  x  =  b'c-  be' ; 
b'c  —  be' 


ab'  -  a'b 


(3). 


ac  —  a  c 
In  like  manner,  y  =  — —      •     •     •     (4). 

By  means  of  formulas  (3)  and  (4)  any  two  simultaneous  equations 
of  the  forms  (1)  and  (2)  may  be  solved. 

Thus,  4x  +  3y  =  31, 

Sx  +  2y  =  22  : 

by  comparison  with  (1)  and  (2), 

a  =  4,     6  =  3/    c  =  31,     a'  =  3,     b'  =  2,     c'  =  22  ; 

by  substitution  in  (3)  and  (4). 

62  —  60  88  —  93       u 


124  APPENDIX. 


EXAMPLES. 

(     -  +  ?   =2     ) 

1.  Given  J     3      4  v.     to  find  x  and  y. 

(  3*  +  4y  =  25   j  j 

By  comparison  with  (1)  and  (2), 

«  =  fc         *   =  *,         c  =    2, 

a'  =  3,         y  ek  4,         c'  s  25  ; 
by  substitution  in. (3)  and  (4), 

]3    —       4  13    ~   4 

(       11a;—    5y  ==  —  1) 

2.  Given  <  >      to  find  #  and  y : 

(-  bx    +  16y  =  124)  J 

by  comparison, 

a=r        11,         b   =  —    5,         c  =  —  1, 
a'  =  -    5,         6'  =       16,         c'  =       124 ; 

oy  substitution, 

_  16    +620  1364  —  5 


176-25  ?        *         176-25 

GENERAL  SOLUTION  OF  THREE  SIMULTANEOUS  EQUATIONS  OF 
THE  FIRST  DEGREE. 

2.  Take  the  equations, 

ax    -f  by    -f-  cz    =  d     •     •     •     (1), 
a'x  -f  b'y  +c'z  =d'    •     ■     .*    (2), 
a"x  4-  b"y  4-  c"«  =  d"  •     •     i     (3). 
From  (1)  and  (2)  we  obtain,  by  eliminating  z, 

(c'a  -  ca')x  4-  (e'A  -  cb')  y  =  c'd  -  cd'     .     .     .     (4). 


ADDITIONAL  EXAMPLES.  125 

In  like  manner,  from  (1)  and  (3), 

(c"a  -  ca")  x  4-  (c"b  -  cb")  y  =  c"d  -  cd"         *     •     *     (5;  ; 

combining  (4)  and  (5)  and  eliminating  y,  we  find 

_  (c"b  -.  cb")  (c'd  -  cd')  -  (c'b  -  cb')  (c"d  -  cd") 
X-{c'a  -  ca!  )  (c"b  —cb")  -  (c"a~  ca")  (c'b  -  c6')       '     '     ™ 

In  like  manner, 

_  (</a  -  ca')  (c"d  -  cd")  -  (c"tf-  ga")  (cW  ~  <Q 
y  -  (^  _  CJ)  [jpl  "Z  cb")  ~  (c"a-  ca")  (c'b  -  cV)        '  '  ™ 


(a"6  -  ab")  (a'd  -  ad')  —  (a'b  —  ah')  (a"d  -  ad") 
(c'a   -  ca')  (6"a-6a")  -  (c"a~*  a"c)  ~(b'a  -  bar ) 


(8). 


Formulas  (6),  (7)  and  (8)  enable  us  to  solve  all  groups  of  simul- 
taneous equations  of  the  form  of  (1),  (2)  and  (3).     Thus, 

2x  -f  3y  +  4z  sa  29, 

Sx  +  2y  +  5s  sa  32, 

4*  4-  3y  4-  2*  =  25 : 

fcy  comparison  with  (1),  (2)  and  (3), 

a    as  2,         6  ■  =s  3,         c    s  4,         rf    =  29, 

a'  =3,         6'  =  2,         c'  =*  6,         *  =  32, 

a"  =  4,         6"*=  3,         c"  =  2,         ^'^25  J 

by  substitution  in  (6),  (7)  and  (8), 

*  1 ( 6  r 12)  (145  ~ 128)  r (15  ~  8)  (58  - 10°) 

(10  -  12)  (     6  -  12  )  -  (  4  -  16)  (15  -  8) 


-  102  +  294  _  192  _ 
12  +  84    ^"90   *" 


126 


APPENDIX. 


Z  = 


[IP  -  12)  (58  -  100)  -  (4  -  16)  (145  -  128) 
(10  -  12)  (  6  -  12  )  -  (4  -    6)  (  15  -  8     ) 

84  +  204  _  288  _ 
12  +  84    ~~  96  ~  °' 

(12  -    6)  (87  -  64)  -  (9  -    4)  (116  -  $0) 
(10  -  12)  (  6  -  12)  -  (4  -  16)  (     4  -  9  ) 

138  -  330 


+  12-60 


192 

48 


1.  Given 


EXAMPLES. 

x  +    y  +  z  =       90 

2a?  —  3y  s=  —  20 

I  2x  —  42  =  —  30  J 


By  comparison  with  (1),  (2)  and  (3), 

a    =  1,         6    =       1,         c    =z       1, 


a'  s=  2,         6'  ss  -  3, 
a"  s=  2,         bn  =       0, 

by  substitution, 

-  4  x  20  -  3  x 


0, 
-4. 


to  find  a?,  y  and  ft. 


rf    =        90 

df  =  -  20, 
rf"  ss  -  30 ; 


&  = 


2.  Given 


-2  x 

-4  +  6  X  3 

-2  X 

—  330  +  6  x  20 

-2  X 

4  +  6x3 

2  x 

200  -  5  X  210 

330      910       „. 

26  ' 


2X-2+6X-5 

x  +     y  +     £  =     6 

«  +  2y  +  3z  =  14 
l&c-y    +  4*=  13 


—  650 

-26 


=  25. 


to  find  a?,  y  and  zi 


ADDITIONAL   EXAMPLES.  127 


by  comparison, 

a    =  1, 

b    = 

1, 

c    =1, 

rf    =    6, 

«'  =  1, 

V    z= 

2, 

c'  =3, 

*  =  14, 

a"  =  3, 

b"=z  - 

1, 

c"  =  4, 

d"  =  13 : 

by  substitution, 

5x4-11     , 
*~2x5-l   -1' 

2x11- 
*  =  2x    5- 

-4 
-1 

-2;     z- 

4X-8+1X5 
2X-4-1 

ELIMINATION  BY  THE  METHOD  OF  ARBITRARY  MULTIPLIERS. 
3.  There  is  a  method  of  elimination  by  means  of  arbitrary  quan- 
tities that  will  often  be  found  useful,  particularly  in  the  higher 
investigations  of  applied  mathematics.  It  consists  in  multiplying 
both  members  of  one  of  the  given  equations  by  an  arbitrary  quan- 
tity, then  adding  the  resulting  equation  to  the  second  of  the  given 
equations,  member  to  member,  after  which  such  a  value  is  to  be 
assigned  to  the  arbitrary  quantity  as  will  reduce  the  co-efficient  of 
the  quantity  to  be  eliminated  to  0. 

To  illustrate,  let  us  take  the  two  simultaneous  equations, 

^  ax  +  by.  =c      ...     (1), 

afx  +  b'y  =  c>     ...     (2); 

multiplying  both  members  of  (1)  by  n,  which  is  entirely  arbitrary, 
we  have 

nax  -+-  nby  ss  nc     •     •     •     (3) ; 

adding  (2)  and  (3),  member  to  member,  and  factoring, 

(na  -f-  af)  x  -f  (nb  -f  b')  y  =  nc  +  c'     •     •     •     (4). 

If  it  be  required  to  eliminate  y,  place 

nb  +  b'  =  0;  .-.     ft  ft*  — —  j 


128  APPENDIX, 

substituting  tnis  in  (4)  and  reducing,  we  find 

—_c  +  e' 

b     +  b'c-  be' 

x  =  ___  _____    ...    (5). 

tf  it  be  required  to  eliminate  x,  place 

a' 
nu  +  «'  ==  0 ;         .*.     n=—  —  ; 

a 

substituting  in  (4)  and  reducing,  we  find  t 

c  4-  C 

a  aer  —  oc 

y&  —-r-—  =  ab'-a'b     '     '     '     «" 

0  -f"  " 

a 
These  values  of  <c  and  y  correspond  to  those  already  deduced  by 
previous  methods. 

As  an  example,  let  it  be  required  to  find  the  values  of  x  and  y 
from  the  equations 

3s-    y^b      ...     (1),       # 
7x  4-  Sy  _=  33     .     •     .     (2)  ; 
multiplying  both  members  of  (1)  by  n, 

3?i„  —  wy  __  5/i     •     •     ♦     (3)  ; 
adding  (2)  and  (3),  member  to  member, 

(3n  +  7)  x  +  (3  -  n)  y  __  5»  +  33     .     •     .     (4) : 
1st.  Assume         3  —  n  __  0 ;         .*.     w  s_  3 ; 
substituting  in  (4),  and  reducing, 
15  +  33 


_=3. 


2d.  Assume         3ra  4-7  =  0  ;         .*.     w=  — 


9  4-7 

7 
3; 


ADDITIONAL  EXAMPLES. 


129 


substituting  in  (4),  and  reducing, 


y^- 


-1+33 


=  4. 


1.  Given 


EXAMPLES. 


s  +  ^y  =  14 


Ux—Qy  =  *    J 


to  find  x  and  y. 


Multiplying  both  members  of  the  first  equation  by  n  and  adding  to  the 
second,  member  by  member, 


making 


n  =  —  -         and  reducing, 

to 


y  =  i2; 

making 

n  =  -        and  reducing, 

x  —  8. 

2.  Given 

■< 

to  find  a;  and  y. 

Reducing  and 

transposing, 

7*-y  =  33     •     •     •     (1), 

IS 

Jy  —  a?  =  19     •     • 

•    (2): 

multiplying  by  n,  and  adding  and  factoring, 
9 


130 


APPENDIX. 


(7w '—  1)  x  -  (n  -  12)  y  d±  33w  +  19 : 


making 

■-> 

we  find       y  = 

:2; 

making 

n  =  12,        we  find        a;  =  5. 

f^+l  =  36] 

' 

3.  Given 

i 

>o               1 

14    6    ,n 

h      to  find  a;  and  y, 

Multiplying  by  n,  adding  and  factoring, 

(9n  +  14)  I  +  (6ti  -n  6)  5  =  36«  +  10 ; 
a;  '  y  , 

nmaking  n  = —-,     we  have     -  =  S;       .  • .  y  =  -, 

9  y  3 


making  n  =  1, 


we  have 


1 


#  = 


MISCELLANEOUS  GROUPS  OF  SIMULTANEOUS  EQUATIONS  OF  THE 
FIRST  DEGREE. 


I.  Given 


j  3z  +  5< 


2 
5y~9 


5a      3y      4  J 


to  find  a?  and  y. 


Combining  and  eliminating  — , 

x 

\25       9/  y  ""45       12' 
whence,  -  =  ttt;         .  • .  y  =  Iff- 


ADDITIONAL  EXAMPLES. 
Combining  and  eliminating  -> 

\9      25/  x  ~  27       12 ; 


131 


whence. 


2.  Given 


1  _  65 
*~~192 


.     *  =  2  $& 

*  65 


+  2y~5 


2a;--  1 


-y  +  1  £=0 


►    to  find  a?  and  y ; 


Clearing  of  fractions  and  reducing, 

*  +  6y==:     15     .     \     .     (1), 
2*-5y  =  -4     .     .     .     (2): 
combining  and  eliminating  ar, 

17y  =  345        ...    yi=2; 
substituting  in  (I),  #  —  3# 


3.  Given                   , 

y-2 

X  —  ~ =  B 

7             ° 
^         *  +10       « 

to  find  x  and  y» 

Clearing  of  fractions  and  reducing, 

7*  -  y  =  33     .     :     .     (i), 

12y  —  x  =  19     .     .     .     (2)  5 

combining  and  eliminating  $ 

83y  =  166;         ...     y  =  2; 

substituting  in  (I), 

a;s:5. 

j$g 

APPENDIX. 

*■ 

[    5\7   4  +  2^24    ] 

4.  Given 

20-2y  +  5,.40f 

to  find  a;  anc 

Clearing  of  fractions  and  reducing, 

hx  +  12y  =  148, 

25s—   2y  =  182; 

combining  and 

eliminating,^, 

155*  =1240;        ,<.     x  te  8 ; 

substituting,  we  find                y  =  9. 

' 

3^3                6 

5.  Given 

- 

?  +  f  =    8  +  2 
ii  +  3  =  10 

to  find  ar,  y  and  *. 

Clearing  of  fractions  and  reducing, 

3of+-2y+    2*:72    •     •     •     (1), 

-ar-f- 3^  +  2*^48     *     .     -     (2), 

& 

c  +           2*  a4  00 

1    * 

•    (8)  i 

combining  (1)  and  (2),  eliminating  y, 

\\x  —Z  =:  120     •     •     • 
combining  (3)  and  (4),  eliminating  zr 

25j  =  300;         .•„    #=12; 
by  substitution  in  (3),  ' z  =  12, 

(1),  y  =  12. 


(4)  J 


ADDITIONAL   EXAMPLES. 


133 


2  +  !  +  7  =  22 

• 

C.  Given 

•  .  y     z 

3  +  1  +  2  =  31 

to  find  x, 

yan»  a 

x  ,  y     z 

Clearing  of  fractions, 

• 

1 

21a:  -f  14y  -f    6z  =  924     . 

■    ■     (1), 

10*  +    6y +15z  =  930     . 

•    •    (2), 

*+    2y  +      2  =  114    . 
combining  (1)  and  (3),  eliminating  *, 

•   •    (3); 

■ 

15a;  -f-  2y  =  240     i 
combining  (2)  and  (3),  eliminating  *, 

••     •     (4); 

5a;  -f  24y  =  780     . 
combining  (4)  and  (5),  eliminating  ar, 

•   •   (5); 

■ji 

70y  =  2100;           ...    $ 
from  (5),                  a:  =  12;     from  (3),     , 

'  =  30; 
=  42. 

-  2ar-  3y  -f  2z  =  13     .     . 

•  (2)  1  t, 

•  (3)  | 

7.  Given     - 

2v  —    x  =  15 

2y+  2=  7 

*  5y  +  3t>  =  32               .     . 

and  v. 

Combining  (2)  and  (4),  eliminating  r, 

10y  +  3ar  =  19     .     . 
combining  (5)  and  (1),  eliminating  ar, 

•    (5); 

2% 

-  62  =  _  1     .     . 

•    (6); 

134: 


APPENDIX. 


combining  (6)  and  (3),  eliminating  z, 

41y  =  41;         .-.     y  =  li 

by  successive  substitutions,        2  =  5,     x  =  3,     v  =  9. 


8.  Given 


a?       y       12      s 

3       5       19      4 

a;  +  2    ' .  24  +  y 

l+^i  +  5 

v.  y       z        8      » -> 


to  find  ft,  y  and  a. 


Transposing,  reducing,  &c, 


2i+3l 

*       y 


4A  =  ± 

2        24 


x  y   x       z       24 

—  5-4-7-  +  5-=—. 
i   x  y         z       24 


(2). 
(3): 


combining  (1)  and  (3),  also  (2)  and  (3),  eliminating  -, 

z 

•     ■     (4), 


_l0I+43l  =  4| 

x  y      24 


el 

1  ";* 

-111™     .     .     . 
y    '24 

(«U 

combining  and  jeliminating  -, 

oq,1       468                     1        2 

234y=  24'         •■•      ^  =  2?     °r     ^=12' 

substituting  in  (5),       x  =  6 ;     whence,     2  =  8. 

f     10+  6y~  4x       4   ] 

9.  Given              < 

6ar—  9y+  3      "  3 

126+  8ar-17y       35 
[  100-12*+  ly  ~~  13  J 

to  find  a;  and  y. 

ADDITIONAL  EXAMPLES.  135 

Clearing  of  fractions  and  reducing, 

-  2a:  +      3y  =  -      1     .     .     .     (1), 
262*-233y  =       931     •     •     .     (2). 
Combining  and  eliminating  x, 

160y  =  800;         .'.     y  =  5; 
by  substitution  in  (1),  x  =  8. 

r  ax  -+-  by  =  c2  \ 
10.  Given  <  a(a  -f  *)__  -   >     to  find  x  and  y. 

Clearing  of  fractions  and  reducing, 

ax  +  by  =  c2     •     •     •     (1), 

ax  —  by  =  b2  —  a2      .     (2) ; 
Sj  addition, 

£2     1     C2  __  G2 

2as  =  62  +  c2-a2;  .'.     *  =  — ^— ; 

by  subtraction, 

a2  4.  C2  _  ^2 


2ty  =  c2  +  «2-  &\  .*•     y  = 


26 


MISCELLANEOUS  EXAMPLES  OF  EQUATIONS  OF  THE  FIRST,  SECOND 
AND  HIGHER  DEGREES,  CONTAINING  BUT  ONE  UNKNOWN 
QUANTITY. 

1.  Given  Sx2  —  4  =  28  -f  A     to  find  x : 
transposing  and  reducing, 

x2  =  16;         .-.     x=  ±4 

2.  Given         ?£±*  -  ?1±*?  =  1 1?  -  5s2,     to  find  *; 


136  APPENDIX. 

Clearing  of  fractions,  transposing  and  reducing, 
x2  =  25  ;         .  • .     s  =  ±  5. 

3.  Given  s2  -f  ab  =  5s2,     to  find  a; ; 

transposing  and  reducing, 

*2  =  V'         '.?     s=±-y^. 

«,  a;   +  7        a;   —  7  7  _ . 

4-Glven        irr^-^4:-^  =  ^rw  wfind*- 

Clearing  of  fractions, 
s*  +  14s3  —  24s2  -  1022s  —  3577  —  s4  +  14s3  -f  24s2  —  1022s 
4-  3577  ==  7s3  -  343s ; 
transposing  and  reducing, 

21s3  =  1701s,     or    s2  ==  81 ;         .  •.     s  =  ±  9. 

/x  —  2  /s  +  2 

5.  Given  V  s~+2  +  V  s~^2  "=  4'     to  find  *> 

multiplying  both  members  by  -y/s  -f  2, 

s  4-  2  / 

V*T^2  4-  -~=  =  4ys  +  2; 
ya;  —  2 


multiplying  both  members  by  yx  —  2. 

s  —  24-s4-2  =  4  </s2  —  4,     or    s  =  2  -/s2  —  4 ; 

squaring  both  members, 

i  /»       i  /»  * 

^  =  4iC2  _  16>     or     X2=  x3;         ...     ajsdbsV^" 

6.  Given  a;  4-  V5x  +  1°  =  8>        to  find  *  J 


ADDITIONAL   EXAMPLES.  f  137 

Transposing  and  squaring  both  members, 

5a;  +  10  =  64-  16*  +  0\ 
whence,  x2  —  21a;  =  —  54  ; 


.       .        ,  21  LA    "   441       21   ,   15 

by  the  rule,         x  =  ~±y>Jb±  +  — =  —±—\ 

.'.     x  =  3,         x  =  18. 
7.  Given  5  %/&  +  7  */x*  =  108,       to  find  x : 

make  3y/x^  =  y  ;     whence,     3^/xi  =  y2  ; 

substituting  and  reducing, 


7  108 


y2  +  e  y  = 


5'  5 


7  /108   '    49  7       47 

whence,  y  =  _  _  ±  ^Z— +  _  =  -  _  ±  _; 


27 

y  =  4     and     y  =  —  —  ; 


__                                        —            /7      27V 
from  which,     a;=±-y/2/3=±8     and    x=±yy3=zk\/ 1 — ) 

8.  Given  3s2  +  10a;  =  57,       to  find  x. 

By  division, 

x2  -f  -— •  x  =  19  ;     whence, 
o 

5  ZtT    ,   25  5       14 

.*.     a;  =  3     and     a;=— 6J« 

9.  Given  (x  -  1)  (x  —  2)  =  1,     to  find  x. 
Performing  indicated  operations  and  reducing, 

x2  —  3a;  =  -  1  ; 


X38  APPENDIX. 


.S^fHsSfflft**^ 


115 

10.  Given  -x2  —  -  x  =  -,     to  find  a:. 

2         '3  8 

Dividing  both  members  by  -,  or  multiplying  by  2, 


-  x  =  -  ;     whence, 


1         /5  .    1       17 

•'•    *=1i,     *=  —  y 

11.  Given 

2* -10      z  +  3      0 

Q            ~              Z           O    =  2'        t0  filld    * 

Clearing  of  fractions, 

2a;2  -  14*  -f  20  -  (5s  +  24  -  a:2)  =  20a:  -  32  -  &r» ; 

A       '  2  39  28  L 

reducing,  a:2  — —  x  = — ;     whence, 


a; 

39 
"10 

v- 

28 
"  5 

+ 

1521 
100  = 

39  ±31 
10       ' 

•  ' 

.     a;  = 

«* 

4 
*  =  5* 

12. 

Given 

1 

1 

1 

o   or  5 

to  find  x. 

Clearing  of  fractions, 

35  (x  +  3  -  x  -f-  1)  =  x2  -f  3*  —  3 ; 
reducing,  x2  -{-  2x  =  143  ;     whence, 

ar  =  -  1  ±  >/Ii4  =  -  1  ±  12; 


ADDITIONAL   EXAMPLES.  139 

.'.      3*9*11,  X=   -13. 

24 

13.  Given  x  -\ —  =  3r  —  4,     to  find  x. 

x-l  % 

Clearing  of  fractons, 

x2  —  x  +  24  =  3a;2  —  3a:  —  4x  +  4 ; 
reducing,  x2  —  3a;  =  10  ;     whence, 


-i*^ 


.     a;  =  5,         a;  =  —  2. 


14.  Given  — — -  H =  — ,     to  find  w. 

x  -f  1  #  6 

Clearing  of  fractions, 

Qx2  +  6a;2  +  12a;  +  6  =  13a;2  +  13a? ; 

reducing,  x2  +  x  =  6  ;     whence, 


.  • .     a;  =  2,         x  =  —  3. 

a;— 4 

15.  Given  —  =  x  —  8,     to  find  x 

2  + y^ 

Since  *    a;  —  4  =  (yaT— 2)   (<v/a7+ 2), 

we  have,  by  performing  indicated  operations, 

•y/aT—  2  =  z  —  8,     or     y^~=  a;  —  6  ; 
squaring  both  merr  bers, 

a;  =  x2  —  12a;  +  36 ; 
or,  x2  —  13a;  =  —  36  ;     whence, 


13  .        /      ~a   r    169       13       5 


140  APPENDIX. 

.-.     x  =  9,         x  =  4. 

16.  Given  17a:2  +  19a;  —  1848  =  0,     to  find  x. 

$ 

O    A      ■  2  J_   19  1848  ^ 

Reducing,  a;2  +  —  x  =  —— ;     whence, 


19         / 


1848        361        —  19  ±  355 


17     '    1156  34 

.-.     a;  =  9  if,     and     *  =:  —  11. 


1  5 

17.  Given  -  x2  -f  -  x  =  27,     to  find  x. 


Multiplying  both  members  by  3, 


a;  =  - 

Given 

15 

4" 

.       X 
X 

-*¥ 

s  =  81; 

whence, 

-  15  ±  39 
4 

13  J- 
-,     to  find 

±01  + 

=  6,     and 

+  4  +  ,/ 

225       - 

16  ~ 

x  =  — 

\ 

18. 

v. 

Transposing  and  reducing, 

/T+4      28 -a;2  ,     ■    ■        28 -a;2 

squaring  both  members  and  clearing  of  fractions, 

a;2  —  16  =  784  —  56a;2  +  x* ; 
reducing,  a;4  —  57a;2  =  —  800  ; 


ADDITION Aju  EXAMPLES,  141 

by  Rule,  Art.  124, 


/w\ /      «™  ,   3249  /I 

8  *  ±  V  T  *  V  -  800  +  -4-  r  *  V " 

hence,  *  =  rh  5,     and     *  s=  zfc  4y2. 


2  ±2; 


Afl  2*-f  9       4*-  3      0   ,  3a:  —  16      ,    -    _ 

19.  Given         _^-  +  — _  ?  3  +-^-,     to  find  », 

Clearing  of  fractions, 

16*2  +  84*  +  54  -h  72*  -  54  =  216*  +  162  + 12*2  -  55*  -  48  5 

reducing,  *2  —  -  *  =  — —  ;     -whence, 


-s-v^ 


14       25       g  dL434 

4    +04~        8       5 


.     *  =  6,     and     *  =ss  — -  4  f « 


20.  Given        *2  +  *  •+■  2  V*2  +  *  -h  4  sa  20,     to  find  *, 
Making  *2  -f-  *  =  y,         and  reducing, 


2  yV  -h  4  =  20  -  y  1 
squaring  both  members, 

4y  +  16  sr  400  —  40y  +  y2  ; 
reducHig,  y2  —  44y  fe  —  384  j       whence, 

y  ==  22  ±  V-384  +  484  s  22  is  10  J 
.-.     ysss32,     and     y  =  12 : 
taking  the  first  value  of  y  and  substituting  in  the  equation, 
*2  4-  *  ~  y, 
*2  -f  *  je  32  $     whence, 


14r2  APPENDIX. 

taking  the  second  value  of  y, 

a;2  -f-  x  =  12 ;     whence, 


21.  «»        yVTi+N/i-i.=,, 

transposing, 

squaring  both  members, 

x t  a;2  -  2  yz2  -~a7  -f  1  -  -1 ; 

whence,  by  reduction, 


.     x  =  3,     *  =  —  4. 
to  find  a:. 


a:2  -  x  -f-  1  =  2  -vA2  -  a-. 
Placing  a;2  — a;  =  ?/     »     *     .     >     (I), 

2/+1  =  2v^ 
squaring  both  members, 

y2  +  2y  -f-  1  =  4y ;     whence, 
y2  -  2y  =  -  1  5         .  • .     y  =  1  ±  -/-  1  +  1,     or     y  =  1  ; 
substituting  in  (1),         x2  —  x  =  1  \ 


i     /rri   i  ±  v^~ 


22.  Given  z3  —  6z  =  6a;  +  28,     to  find  £ : 

transposing,  a:2  —  1 2a;  =  28  ;     whence, 

x  e=  6  ±  -y/28  -f-  36"=  6  =b  8  j         .->     »  =  14,     a;=«~2» 


ADDITIONAL  EXAMPLES.  143 

23.  Given  x*n  —  2x3«  -f  x*  —  6  =  0,     to  find  <b  : 

making,  xn  =s  y  ;     whence, 

y*  -  2y3  +  y  -  6  =  0 ; 
causing  the  second  term  to  disappear  (Arts.  263  and  313), 

3     _01 

By  the  rule  for  solving  trinomial  equations  (Art.  124). 


/3  /91    ,     9  /3±10       1     /o  -  1A 

...     2=±ivT3,     and     ^±1^=17 

,                      1    .                                1±#"        i             Idb^/'^f 
tut,         y  =  -  +  z;         ./•     y= f— ,     and     y  = 1 


, —        .         n  /l  ±yT3"        n  /I 


_±i/-7 

also, 


2        '  V  2 


24.  Given  „   |     '■    ^      rtf  +  a;  +  8,     to  find  ar. 


Clearing  of  fractions, 

i*  +  2«3  +  8  =  2*  +  2a;3  +  3a:2  -f  2s  —  48 ; 

reducing,  a;2  -f-  r  #  as  —  ;     whence, 


•1         M  ,  1      —  1  ±  13 

1:=-3±VT  +  9==— 3—  5 

.  • .     a;  =  4,     and     a:  =  —  4 J. 

2 
25.  Given  x2  —  1  =  2  +  -,     to  find  a?. 

a; 

Reducing,  s3  —  3a;  —  2  =  0     •     ■     •     (1)  ; 


144  APPENDIX, 

comparing  with  xa  -f-  px  -f-  q  =  0, 

by  Cardan's  formula, 

x  =  yr  +  \/r  =  2  s 

dividing  both  members  of  (1)  by  x  —  2, 

a:2  -h  2#  +  1  =  0  j     whence, 

>v  a;  =  —  1, 

and  the  two  roots  are  each  equal  to  —  1  ;  hence,  the  three  roots 
aro  -f  2,  — >  I,  and  —  1. 

26.  Given  2a;2  -f  34  ==  20a;  4  2,     to  find  x. 
Transposing  and  reducing, 

a;2-  10a;  £=  —  16;       .*•.     x  m  8,     x  =  2. 

27.  Given  a;*  =  56s  8  +  a:6,     to  find  x. 

multiplying  both  members  by  x  ,  and  reducing, 

**-**  =  56: 
comparing  with  trinomial  equations  (Art.  124),  we  find 


3  ,     1       2 

n^-,     and     -  =--  -  ; 


hence,  by  rule, 


taking  the  upper  sign,         x  ss  8    =4, 

"    lower     "  x  =  (-7)3"  =  ^/49. 

28.  Given         x3  —  12s2  -+-  4a;  -f  207  =  0,     to  find  & 


ADDITIONAL   EXAMPLES. 


145 


A  superior  limit  of  the  positive  roots  is  13  (Art.  284) ;  a  supe- 
rior limit  of  the  negative  roots  (numerically),  is  —  7  (Art.  286). 
By  the  method  of  Art.  297,  rejecting  -f  1  and  —  1,  which  are 
not  roots,  we  find 

Divisors, 


9, 

3, 

-    3 

23, 

69, 

-69 

27, 

73, 

-65 

3, 

-9, 

-1, 

0; 

hence,  9  is  a  commensurable  root. 

Dividing  both  members  by  x  —  9,  we  find 

X2  _  %x  _  23  =  0  ;     or,     x2  —  Sx  =  23  ; 


3 

y/28  + 

93  ±  yloT 

4~          2 

29.  Given            x3  +  3x2  —  Qx  - 

-8  =  0,     to 

find  x. 

This  is  solved  in  a  manner  similar  to  the  preceding. 

A  superior  limit  of  positive  roots  is  4,  and  of  negative  roots 

(numerically),   —  7. 

Divisors            4,           2, 

h 

-  1,      -  2, 

-4, 

-2,     -    4, 

-8, 

+  8,     +4, 

+  2, 

-8,     -10, 

-14, 

+  2,     -2, 

-4, 

-2,     -    5, 

-14, 

-2,     +1, 

+  1, 

+  1,     -    2, 

-11, 

+  1,     +4, 

+  4, 

-    1, 

-11, 

-1,     -2, 

-1, 

0, 

-10, 

0,     -1, 

0; 

hence,                    x  =  2, 

X  =z   — 

1,        *=- 

-4. 

10 

!^>  /jp^s^ 

**/*  Of  THB 

rTJ2ri7BE 


146  APPENDIX. 

30.  Given  x3  -f  9x  —  1430  =  0,     to  find  x. 

By  the  same  rule  as  before,  we  find  14  for  a  superior  limit  of  the 
real  positive  roots,  and  from  Art.  288  we  see  that  the  equation  has 
no  real  negative  roots.     By  the  rule  (Art.  297),  we  have 

13,  11,  10,  5,  2,  1, 

-  143,     -  286,     -  715,     -  1430, 

-  134,     -  277,     -  706,     -  1421, 


-110, 

-130, 

-101, 

-121, 

-   8, 

-  U, 

,..., 

-   1, 

..., 

o, 

353, 

-  1421, 

..., 

-  1421, 

..., 

-  1420, 

ihence,  1 1  is  the  only  commensurable  root. 
Dividing  both  members  by  x  —  11,  we  have 

x2  +  11*  =  -  130; 


11  -l.     /     10A  ,   121       -  1  dfc  -/-  399 

no 

31.  Given  Vx  +  Vx  +  7  = >     to  find  x. 

'Clearing  of  fractions  and  transposing, 


-vA2  +  7*  =  21  -x) 
squaring  both  members  and  reducing, 

49*  =  441;         .*.     x  —  9. 


32.  Given  -y/a  -f  x  —  y/a  —x  sa  -\/ax,     to  find  x. 

Squaring  both  members  and  reducing, 


2a  —  ax  =  2  4/ a2  —  *2 ; 
squaring  both  members, 

4a2  —  4a2*  +  a2*2  =  4a2  —  4**  ; 


ADDITIONAL  EXAMPLES.  147 

reducing  and  dividing  both  members  by  a;, 

*(«2  +  4)=4a»;         .'.     *  =  — j 


33.  Given       '  >y/4a  f-  c  ss  2  y^6  -f-  a:  —  -y/^     to  find  r. 
Squaring  both  members  and  reducing, 


(a  —  b)  —  x  =  —  y^a:  -f-  a;2  j 
squaring  both  members, 

hence,  (2a  —  6)  x  sa  (a  —  6)2  ;         .  • .     x  ss  -^ £-• 

34.  Given       y"4a  -f  C  -f  -/"  +  *  *=  2  V*       2a>     to  ^n^  *• 

Squaring  both  members  and  reducing, 

2  -v/4a2  -f  5oa;  -f-  a;2  s2«-  13a ; 
squaring  both  members, 

16a2  4-  20a*  +  4a;2  =  4*2  —  52a*  +  169a2 ; 

1  iy 

reducing,  72ax  ss  153a2;         »*»     «  =  -— • 

o 

35>  Given  (TT$  +  (T^~$  =  "'   t0  find  * 

Dividing  both  terms  of  the  first  fraction  by  (1  4-  #),  an(*  of  thd 
se<fmd  by  1  —  x,  we  have  9 

\  —x  +  x**  ,    l^s-fs2 
1 +«  1  —S  * 

clearing  of  fractions, 

1  —  2a:  +  2a;2  -  x3  4-  1  -f  2a;  -f  2a:2  -f  x3  ss  a  —  az2 ; 

/a 2 

reducing,         (4  -f-  «)  £2  =  a  —  2 ;         » • .     x  =  =fc  w  * 


148  APPENDIX. 


MISCELLANEOUS  EXAMPLES   OF  SIMULTANEOUS  EQUATIONS    Of 
THE  SECOND  AND  HIGHER  DEGREES. 

iz3  +  y3  =  ]Sxy     .     .     .     (1)  )     * 

1.  Given       \  y  y  w  I    to  find  x  and  y. 

\  x   +  y    z=z  12         .     ,     .     (2)  I 

Make  z  =  v  +  w,     and    y  =  v  —  w. 

From  (2),  we  have     (v  -{-  w)  +  (v  —  tv)  =  12 ;         .'.     v  =2  0. 

From  (1),  we  have,     (v  +  w)3  +  (v  —  w)3  =  18  (v2  —  w2)  ; 

or,  reducing,  v3  +  3vw2  =  9  (v2  —  w2) ; 

substituting  the  value  of  v,  we  have 

216  +  18w2  s=  9  (86  -  w2),     or    27w2  =  108  ;         .  • .     w  =  dt  2  , 

hence, 

*  =  i;  +  w  =  6±2i8     and     4;  y  =  v  —  «/=:  6+2  =4    and    8. 

(*2  +  2/2  =  53     .     .     .     (i)j 

2.  Given  ■{  }•   to  find  x  and  y. 

(  zy  =  14     .     .     .      (2)) 

Multiplying  both  members  of  (2)  by  2,  and  adding  and  subtracting, 

we  have 

x2  +  2xy  +  y2  =  $l',         .'.     x  +  y  -  ±  9, 

z2  —  2#y  -f  y3  —  25  ;         . ' .     a;  —  y  =  ±  5  ;  f 

hence,         a:  5=  -f-  7,     and      —  7     y  =  +  2,     and     —  7. 

I 
(**  +  y*  =  82     .     .     .     (1)) 

3.  Given  ]  Y   to  find  ar  and  y. 

(x  +y   =    4     .     .     .     (2)) 

Raising  both  members  of  (2)  to  the  4th  power,  adding  to  (1),  mem- 
ber to  member,  and  dividing  by  2, 

x4  -f  2z3y  4-  ox2y2  -f  2#y3  -f  y*  as  169  j 


ADDITIONAL   EXAMPLES.  149 

extracting  the  square  root  of  both  members, 

x2  +  xy  +  y2=.\z    .    .    .    (3); 

squaring  both  members  of  (1), 

x2+2xy  +  y2  =  16  .     .     .     (4); 

subtracting  (3)  from  (4),  member  from  member, 

xy  =  S,     or     Sxy  =  9    .     .     (5); 

subtracting  from  (3),  member  from  member, 

x2  —  2xy  4-  y2  =  4  ; 

whence,  x  —  y  =  ±  2     .     .     .     (6). 

Combining  (2)  and  (6), 

x  —  3,     and     1 ;         y  =  1,     and     3. 

j  5*   -f  3y   =  19     .     .     (1)  ) 
4.  Given  <  •    >•   to  find « and «, 

(7z2-2y2  =  10     .     .     (2))  y 

From  (1),  we  find 

_  19  —  3y  _  361  -  114y  +  9y2 

X~        5~~;  '•     f    -  ~25~ 

Substituting  in  (2),  and  reducing, 

2_798_    _22T7# 
V         13  ~         13    ; 


399^      /— 2277  ,    159201 

whence,  y  =  _  ±x/_^_  +  — _  = 

759 

hence,  y  =  — ,     and  y  =  3  ; 

,     .      .  406 

by  substitution,       x  = — -,     and  a:  =  2. 

lo 


399  ±  360 
13        5 


150  APPENDIX. 

Ix   +  4y  =  14     .     .     (1)) 

5.  Gri*\>*i  S  >  to  find  a?  and  y. 

(y2  +  4z  =  2y+ll     (2)) 

From  (1),         v  =  14  —  4y,     or    4a:  =  56  —  16y ; 

subtracting  ana'  reducing, 

y2  _  I8y  =  _-  45 ; 


whence,    y  =  0  db  y—  45  +  81  =  9  ±  6  =  15     and    3 ; 
hence,  x  =  2     and     —  46. 

(  s2  +  4y2  =  256  -  4ary     .     .     (1)) 
6.  Given        \  \  to  find  x  and  w. 

(  3^ -a:2  =  39     ...     (2))  * 

Transposing  in  (1),  and  extracting  the  square  root  of  both  members* 

a  +  2y=±16;         .-.     a:=±I6-2y; 
or,  *'  =  256=F64y  +  4y2; 

substituting  in  (2),  and  reducing, 

y2  rp  64y  =  -  295  ; 


whence,  y  =  ±  32  ±  -y/  -  295  +  1024  =  ±  32  ±  27 ; 

.  * .  y  =  ±  59  and  ±  5  ; 

substituting,  x  =  db  102     and     ±  6. 

(a:2-    y2  =  24     .     .     (1)) 
7.  Given  ■{  *  )•  to  find  ar  and  y. 

(z2  +  a:y   =  84     .     .     (2)) 

Subtracting  (1)  from  (2),  member  from  member, 
y2-f*y  =  60     .     .     .     (3); 
adding  (3)  to  (2),  member  to  member, 

x2  +  2xy+y2  =  U4;         .  \     x  +  y  =  db  12    .     .     .     (4). 


ADDITIONAL   EXAMPLES. 


151 


Dividing  (1)  by  (4),  member  by  member, 

x-y  =  ±2; 
hence,  x  =  db  7,        y  =  ±  5. 

-y=  4   .   •   .    (l) 

*y  =  45     .     .     .     (2) ) 

x  =  y  -f  4,     which,  in  (2),  gives 


8.  Given 


' 


to  find** 


From  (1), 


y2  +  4y  =  45;         ...     y  =  -2±v/45TT=  -2  ±7; 
.  • .     y  =  -f  5,     and     —  9  ; 
and  by  substitution,        x  =  +  9,     and     —  4. 

W  +  *y2  =  i2  •   •   •   0) 

9.  Given  i 

(   *  +  ay3=18     .     .     .     (2) 

Dividing  (2)  by  (1),  member  by  member,  and  reducing, 


y   to  find  x  and  y. 


1  +y3  _  3  1  -  y  +  V2  _  3 

y(i+y)~2'   or         y        _2; 


y2-2y=~1 


by  reduction, 

whence,  y=4±V— 1  +  16 


by  substitution, 


10.  Given 


.     y  =  2,     and     y  =  - ; 
x  =  2,     and     a?  =  16. 


4       4' 
1 


1         1 

—   ••  —   =  a     . 

x        y 


L «"     jr 


(2)  J 


)-   to  find  x  and  y. 


152  APPENDIX. 

From  (1),  we  find, 

1  1  1  .      2a      1 

-  =  a ,     or    —  =  a2 1-  — ; 

y  x  y2  x    r  x* ' 

substituting  in  (2),  and  reducing, 

I  —  a  -h    ih  1  a2     a2     a  ±  v^"--^ 
■'""s~"2"~V     2      +4  ~  2  ; 


by  substitution,  -   = — • 

y  2 

2 


hence,  #  = ,     and  y  =  — 

a  ±  -v/26  -  a2  a  =f  V26  -  "2 

11.  Given  J  y2       y        9  v  '  I   to  find « and  y. 

(   *  -  y  =  2         •  .     (2)  j 
Clearing  (1)  of  fractions, 

9s2  +  S6xy  =±  85y2     .  .     .     (3). 
From  (2),         a;  =  2  +  y ;         .'.    a2  =  4  +  4y  -f  ya ; 
substituting  in  (3),  and  reducing, 


r2  — 


27  9 


y*-TRy  = 


10  J       10 


27  /9    ,   729       27  ±33 

whence,  y  =-  ±  yj-  +  —  =  __; 

17 

by  substitution,  x  =  5     and    £  =  —  • 


ADDITIONAL   EXAMPLES.  153 

12.  Given     J  y2    -  x2  T  y  T  a;        4  w  V  to  find  a;  and  y. 


*  -  y  =  2 (2) 

y   '    x  " 


x        II 

Make         -  -\-  -  =  z    (3) ;     whence,  by  squaring,  &c, 


y2       P 
hence,  from  (1),  by  substitution  and  reduction, 


z2  +  z 


_ 35  l  /35       1  _  -  1  dbft 


5      ,,       7 

or,  ^  =  ^     and     --; 

substituting  the  positive  value  of  z  in  (3),  and  clearing  of  fractions, 

5 

x2  +  y2  =  -xy     •     •     •     (4), 

From  (2),         x  =  y  +  2 ;         .  • .     a2  =  y2  -f  4y  +  4, 
and  #y  =  y2  -f  2y  ; 

substituting  in  (4)  and  reducing, 


y2  +  2y  =  8;         .'.     y  =  -  1  ±  V/8+T=  -  1  ±  3 ; 
hence,  y  =  2     and     y  =  —  4 ; 

by  substitution,         #  =  4     and     y  =  —  2. 

r  *2  +  y2 -f.  z2  =  84     •     •     -     (1)  N 
13.  Given  -j    a?   +  y   -f  *   =  14     •     •     •     (2)    I     to  find  ar,  y, 
I  xz  =  y2  ...     (3)  J  and  * 

Substituting  in  (1)  and  (2)  the  value  of  y  from  (3),  and  reducing 


x2  +  2arz  -f  z2  =  84  +  rz     or     x  +  *  =  -y/84  -f-  ars     •     .    .  (4). 
From  (2)  and  (3),  x  +  z  _  14  _  y^"    .     .     (5). 


154  APPENDIX. 

Equating  the  second  members, 

14  -  i/xz—  -/84  -f  xz ; 
squaring  both  membeis, 

196  -  28  i/x~z  -{-  xz  =  84  +  xz ; 
•educing,  -yfxz  _-  4,     or    xz  =  16     •     •     •     (6) ; 

hence,  from  (5),  x  +  z  =  10     •     •     •     (7) ; 

1  (\ 

aobstituting  in  (7)  for  z  its  value  — >  and  reducing, 

x*  -  10*  =  -  16 ; 


^.httfee,  x  —  5  ±  -/—  16  +  25  =  5  ±  3 ; 

or,  a;  =  8,         a:  a=  2 : 

by  substitution,     z  =  2,     2  =  8,     and     y  ==  ±  4. 


14.  Given     ^  •       ,  '  f.   to  find  x  and  y. 

(  x2  .+  y2  =  41   •  •  (2)  ) 


From  (1),  by  transposition, 


V*  +  y  =  12  -  (*  +  y) ; 
squaring  both  members, 

x  +  y  =  144  -  24  (a?  +  y)  +  (*  +  r)»j 
reducing,  (a;  +  y)2  —  25  (x  -f-  y)  =  —  144 ; 


,    25  /      7777*^5        ,   25  ±7 

...    iC4.y  =  +_±v/_i44  +  — =+-^— ; 

whence,  a:  -f  y  =  16,     or     a-  -f  y  =a  9. 

Tlie  first  value  does  not  satisfy  (1),  unless  the  radical   have   the 
negative  sign  ;   adopting,  therefore,  the  second  value,  from  which 

x  =  9  —  y,     or     x2  =  81  —  18y  +  y2, 


• 

ADDITIONAL   EXAMPLES. 

which 

in  (2) 

gives, 

after  reduction, 

w- 

y 

2  -  9y  =  -  20 ; 

whence, 

9 
y  =  2 

-20  + 

81       9±1 
4  -      2      ' 

.*.    y  = 

by  substitution, 

x  =  4     and     a; 

=  5. 

15. 

Given 

ir 

-.Jf  m      3     •     • 

•     0) 
.     (2) 

155 


to  find  x  and  y. 


Cubing  both  members  of  (2),  subtracting  from  (1),  member  from 
member,  and  dividing  both  members  by  3,  we  have 

x2y  —  xy2  =  30,     or     (.*  —  y)  xy  =  30     •     «     •     (3)  ; 

dividing  (3)  by  (2),  member  by  member,  -" 

1A  .  10 

*y  =  10;  .'.     y  =  — ; 

substituting  in  (2),  and  reducing, 

x2  -Sx  =  10: 


3  /  9       3       7 

whence,  x  =  -  ± yj  10  +  -  =  -  ±  -; 

.  • .     a;  =  5,         x  =  —  2  ; 
by  substitution,  y  =  2,     and     3/  =  —  5. 

f*»+'f*9/iyr=    208     •     •     (1)  ) 

16.  Given     I  [  to  find  x  and  y, 

(y*  +  y^/*V  =  i053    •    •     (2)  )       * 

These  equations  may  be  written, 

**+*V=    208,     or     **(**+y*)=    208     •     .     .     (3), 


y*  +  y***  =  1053,  y*(y* +**)  =  1053     -     - 


(4> 


156  APPENDIX. 

Dividing  (3)  by  (4),  member  by  member, 
**        208        16 


i-       1053       81 

y 

extracting  the  4th  root  of  both  members, 

y 


(5); 


substituting  in  (3), 


A/1  1 A 

y2  4-  77T  V2  —  %®8  ;     whence, 


20S 

—  y2  =  208,     or    j/  =  729;         ...     ^^±27, 

and  by  substitution,  x  =  ±    8. 

MISCELLANEOUS   PROBLEMS. 

1.  A  courier  starts  from  a  place  and  travels  at  the  rate  of  4  miles 
per  hour ;  a  second  courier  starts  after  him,  an  hour  and  a  half 
later,  and  travels  at  the  rate  of  5  miles  per  hour :  in  how  long  a 
time  will  the  second  overtake  the  first,  and  how  far  will  he  travel  ? 

Let       x     denote  the  number  of  hours  travelled  by  2d  courier  : 
then  will  x  +  1  \     " 

5x 
and       4(x  +  1^)    " 
From  the  conditions  of  the  problem, 

5x  -—  4  (x  +  11)  ;         .  • .     x  =  6     and     hx  =  30. 

2.  A  person  buys  4  houses  for  $8000 ;  for  the  second  he  gave 
half  as  much  again  as  for  the  first ;  for  the  third,  half  as  much  again 
as  for  the  second ;  and  for  the  fourth,  as  much  as  for  the  first  and 
third  together  :  what  does  he  give  for  each  ? 


(( 

a 

1st 

"  miles 

a 

2d 

u       a 

u 

1st 

ADDITIONAL  EXAMPLES.  157 

Let     x  denote  the  amount  paid  for  1st  house  :  then  will 

x  +  %         "       "         "  "      «    2d       " 

2x  +  ~         "       "         «  "      "    3d       * 

4 

Sx  -f  %  "       "         "  "      "    4th      " 

4 

From  the  conditions  of  the  problem, 

8* -=8000;  .\     *  =  1000 

x  +  *  =  1500 

2x  -f  S  .-.=  2250 
4 

3*  -f  ?  =  3250 
4 

3.  A  and  B  engaged  in  play :  after  A  had  lost  $20,  he  had  one 
third  as  much  as  B  ;  but  continuing  to  play,  he  won  back  his  $20, 
together  with  $50  more,  and  he  then  found  that  he  had  half  as  much 
again  as  B  :  with  what  sums  did  they  begin  ? 

Let  x  and  y  denote  the  sums  with  which  A  and  B  began. 

Then  from  the  conditions, 

™       V  +  20 


x  +  50  =  (y  -  50)  x  1 J 


Whence,         Zx  -~    60  as    y  +    20 ;         . '.     a:  =    »70 
2a;  rf  100  =  Zy  -  150  ;  y  =  130. 

4.  A  can  do  a  piece  of  worl    in  10  days,  which  A  and  B  together 
can  do  in  7  days ;  in  how  many  days  can  B  do  it  alone  ? 
Let  x  denote  the  number  of  days. 


158  APPENDIX. 

Since  A  and  B  together  can  do  -J-  of  the  w  t>rk  in  1  day,  A  can  do 

y\  of  it,  and  B,  -  of  it  in  1  day  ;  hence,  from  the  relations  existing, 
x 

10       x       l\  a;       70'  5 

5.  A  person  has  $650  invested  in  two  parts  t  the  first  part  draws 
interest  at  3  per  cent,  and  the  second  at  3. J-  per  cent,  and  his  total 
income  is  $20  per  annum  :  how  much  has  he  invested  at  each  rate  ? 

Let  x  denote  the  number  of  dollars  at  3  per  cent :    then  will 
650  —  x  denote  the  number  at  3£» 
From  the  conditions, 

3x       050  -*      „,  _20. 

whence,        Sx  +  2275  —  3£  c  sb  2000 ; 

.  •.  5  s  275,  or  *  =  550 ;    .  • .  650  -  x  =  100. 
2 

6.  A  boatman  rows  with  the  tide,  in  the  channel,  18  miles  in  1-J 
hours ;  he  rows  near  the  shore  against  the  tide,  which  is  then  only 
three-fifths  as  strong  as  in  the  channel,  18  miles  in  2J  hours  :  what 
is  the  velocity  of  the  tide  per  hour  in  the  channel  t 

Let     x     denote  the  velocity  of  the  tide  in  the  channel : 

Sx 

then,    -r-  "      "  "         "  "         near  shore ; 

5 

and      1 18  —  —J  -r  1?  will  denote  the  rate  of  rowing,  neglecting  tide 

also,     (lS  +  ^-f-2*     "        «  «  «        f 

/10       Sx\       2       /,.   ,   27*\       4. 
henee,  (l8  -  -)  X  -  -  (l8  +  w)  X  gi 

or,  12  -x  =  8  +  ^5        ;*,'_'•**•§. 


ADDITIONAL  EXAMPLES.  159 

7.  A  garrison  had  provisions  for  30  months,  but  at  the  end  of  4 
months  the  number  of  troops  was  doubled,  and  3  months  afterwards 
it  was  reinforced  by  400  troops  more,  and  the  provisions  were  ex- 
hausted in  15  months  :  how  many  troops  were  there  in  the  garrison 
at  first  1 

Let  x  denote  the  number  of  men  at  first ;  then  will  30#  denote 
the  number  of  months  that  one  person  could  subsist  on  the  provi* 
sions,  or  the  number  of  month",  y  rations  in  the  garrison. 

4x     denotes  the  number  of  monthly  rations  used  in  4  months, 
Gx        "  "  "  "  "    the  next  3       " 

(2*+400)8  "  "  "  "  "         "        8      " 

hence,       2Gx  +  3200  =  30#,     or    4x  m  3200  ,       .  i .  x  tc  800. 

8.  What  is  the  number  whose  square  exceeds  the  number  itself 
by  6? 

Let  x  denote  the  number. 


From  the  conditions, 


1  ±5 


.  • .     x  2=  3     and     —  2. 

9.  Find  two  numbers  such  that  their  sum  shall  be  15,  and  the  sum 
of  their  squares  117. 

Let  x  and  y  denote  the  numbers. 
From  the  conditions  of  the  problem, 

x  +y    =    15     •     •     •     (1), 

x2  +  y2  =  117     •     i     .     (2). 
From  (1)         x  sbe  15  —  y,     or     x2  z-  225  —  30y  +  y2 } 


160  APPENDIX. 

substituting  in  (2)  and  reducing, 

y*  -  15y  -  —  54, 

15  ^      /      7A    ,    225       15       3  „ 

y=-2-±y-54  +  -r  =  y±-,     or     y  =  9     and     *  *x  6, 

or     a;  ss  9     and     y  =  6. 

10.  A  cask  whose  contents  is  20  gallons,  is  filled  with  brandy  ;  a 
certain  quantity  is  drawn  off  into  another  cask  of  the  same  size, 
after  which  the  latter  is  filled  with  water  :  the  first  cask  is  then,  filled 
with  this  mixture ;  it  then  appears  that  if  6J  gallons  of  this  mixture 
be  drawn  from  the  first  into  the  second  cask,  there  will  be  equal 
quantities  of  brandy  in  each.  How  much  brandy  was  first  drawn 
off? 

Let  x  denote  the  number  of  gallons  first  drawn  off.  Then  will 
20  —  x  denote  the  quantity  remaining  as  well  as  the  quantity  of 

x 
water  added  to  the  second  cask  ;  — j  will  denote   the  quantity  of 

brandy  in  each  gallon  of  the  mixture,  and 

x  x2 

*X20'     °r    20 

will  denote  the  quantity  of  brandy  returned  to  the  first  cask,  which 
will,  therefore,  contain 

gallons  of  brandy.     Each  gallon  of  this  new  mixture  will  contain 
jfc  of  the  brandy  in  the  cask,  or 

400  —  20a;  +  *2  . 
400  ' 

hence,  6f  gallons  will  contain 

400  —  20*  -f  a?8 
60 


ADDITIONAL  EXAMPLES.  161 

gallons ;  and  after  this  is  drawn  off,  10  gallons  must  remain ;  hence, 
400  —  20a?  -f  x2      400  —  20*  -f-  x2 


20  60 

whence,  800  —  40*  -f-  2s2  =  600, 

or,  z2  —  20*  =  100; 


=  10; 


* 


=  10  ±  v-HW+Too,    or    *  =  10. 


11.  What  number  added  to  its  square  will  produce  42? 
Let  x  denote  the  number. 

From  the  conditions  of  the  problem, 

x2  +  *  =  42  ; 

1  ±  13 


•.     af=  — -J+  V42  +  }  = » ;    .*.  *  =  6  and  *  =  ~7. 

lit 

12.  The  difference  of  two  numbers  is  9,  and  their  sum  multiplied 
by  the  greater  gives  266  :  what  are  the  two  numbers  ? 
Let  *  and  y  denote  the  numbers. 

From  the  conditions  of  the  problem, 

******     •     •     •     (1), 

*  (*  +  y)  =  266     •     (2). 

From  (1),  y  =  *  —  9  ;         substituting  in  (2), 

*  (2*  —  9)  =  266,     or     x2  —  -  x  =  133  ; 

,  9  '        /         "    81       9  ±  47 

whence,  £=-±^133  +  -  =  —— ; 

.*.     *  =  14,         *  =  — 9£; 
whence,  y  =  5,         y  =  —  18^. 

13.  A  person  travelled  105  miles :  if  he  had  travelled  2  miles 

11 


162  APPENDIX. 

per  hour  slower,  he  would  have  been  6  hours  longer  in  completing 
the  journey  :  how  many  miles  did  he  travel  per  hour  1 

Let  *  denote  the  number  of  miles  travelled  per  hour.     Then  will 

denote  the  number  of  hours. 

* 

From  the  conditions, 
105         105 


+  6,     or     105*  =  105*  -  210  +  6*2  -  12* 


x  —2         x 
reducing,  *2  — •  2x  =  35  ; 

.-.     *  =  1  ±  V35  +  1  =  1  ±  6  ;         .-.     x  =  7. 

14.  The  continued  product  of  four  consecutive  numbers  is  3024  : 
what  are  the  numbers  % 

Let  *  denote  the  least  number. 

From  the  conditions  of  the  problem. 

*(*  +1)  (*  -f-  2)  (x  +  3)  =  3024, 
or  *4  -f  6*3  +  ll*2  -f-  6*  —  3024  ±s  0. 

A  superior  limit  of  the  real  positive  roots  is  9  (Art.  284).  Ne- 
glecting the  divisor  1,  and  all  negative  divisors,  we  may  proceed  by 
the  rule  (Art.  297),  as  follows  : 


9, 

8, 

% 

6, 

4, 

3, 

2, 

-336, 

-378, 

-432, 

-504, 

-756, 

-  1008, 

-  1512, 

-330, 

-372, 

-  426, 

-498, 

-750, 

-  1002, 

-  1506, 

"j 

**> 

"? 

-  83, 

**? 

-  334, 

-  753, 

~» 

**» 

"i 

-  72, 

i 

-  323, 

-  742, 

••» 

'*» 

*■? 

-  12, 

'*> 

"i 

-  371, 

"» 

"i 

*'j 

-   6, 

"j 

**> 

-  365, 

"» 

"i 

"*> 

-  h 

**» 

"•. 

**» 

"»  ",  ••>      —     0,         .-,  .-, 

Hence,  6  is  the  required  value  of  *,  and  the  numbers  a?e  6,  7,  8 
and  9. 


ADDITIONAL   EXAMPLES.  163 

15.  Two  couriers  start  at  the  same  instant  for  a  point  39  miles 
distant ;  the  second  travels  a  quarter  of  a  mile  per  hour  faster  than 
the  first,  and  reaches  the  point  one  hour  ahead  of  him :  at  what 
rates  do  they  travel  ? 

Let  x  denote  the  number  of  miles  per  hour  of  first  courier. 

39 

Then  will  —  denote  the  number  of  hours  he  travels. 
x 

From  the  conditions, 

x  x  -+-  -J-  4  4 

reducing,  x7,  -f  -  x  =  -— - ; 


1  J.      /39   .     1        -  1  ±  25 

16.  The  fore-wheels  of  a  wagon  are  5 J-  feet,  and  the  hind-wbeek 
7£  feet  in  circumference ;  after  a  certain  journey,  it  is  found  that  the 
fore- wheels  have  made  2000  revolutions  more  than  the  hind-wheels : 
how  far  did  the  wagon  travel  ? 

Let  x  denote  the  number  of  feet. 

From  the  conditions  of  the  problem, . 

'      .   4-^  =  2000; 

1197 
multiplying  both  members  by    -55-, 


Tfir- 

-5i* 

2394000 
32 

598500 

8  ' 

57  * 

-42  a; 

=  598500, 

15x 

=  598500, 

X 

=  39900. 

164  APPENDIX. 

17.  A  wine  merchant  has  2  kinds  of  wine ;  the  one  costs  9  shil- 
lings per  gallon,  and  the  other  5.  He  wishes  to  mix  them  together 
in  such  quantities  that  he  may  have  50  gallons  of  the  mixture,  and 
so  that  each  gallon  of  the  mixture  shall  cost  8  shillings. 

Let  x  and  y  denote  the  number  of  gallons  of  each,  respectively. 

From  the  conditions, 

^+  y  =  50    .    .    .    .    (1), 

9*  +  by  =  $(x  +  y)     .     .     (2); 
substituting  for  x  -f  y  its  value  in  (2), 

9z  +  5y  =  400    .     .     .     .     (3); 
combining  (1)  and  (3), 

Ay  =  50 ;         .  •.     y=  12-J,     and     x  =  37£. 

*18.  A  owes  $1200  and  B,  $2500,  but  neither  has  enough  to  pay 
his  debts.  Says  A  to  B,  "Lend  me  the  eighth  part  of  your  fortune, 
and  I  can  pay  my  debts."  Says  B  to  A,  "  Lend  me  the  ninth  part 
of  your  fortune,  and  I  can  pay  mine :"  what  fortune  had  each  % 

Let  x  and  y  denote  the  number  of  dollars  in  the  fortunes  of  A 
and  B. 

From  the  conditions  of  the  problem, 

x  +  1  =  1200,  x     or         &r  +   f  =2    0600, 

8 

y  +  ^  ~  2500,         or  x  -fify  =  22500  j 

9 

combining  and  eliminating  ar, 

71y  =  170400  \         .'.     y^  2400,         x  =  900. 

19.  A  person  has  two  kinds  of  goodsr  8  pounds  of  the  first,  and 

9  of  the  second,  cost  together  $18,46;  20  pounds  of  the  first,  and 

16  of  the  second,  cost  together  $36,40 :  how  much  does  each  cost 

per  pound  % 


ADDITIONAL   EXAMPLES.  165 

Let  x  and  y  denote  the  cost  of  a  pound  of  each  in  cents. 
From  the  conditions  of  the  problem, 
8*  +    9y  =  1846, 
20*  +  16y  =  3640  ; 
combining  and  eliminating  x, 

ISy  =  1950  :         .  •.     y  =  150,     and    x  =  62. 

20.  What  fraction  is  that  to  the  numerator  of  which  if  1  be 
added  the  result  will  be  J,  but  if  1  be  added  to  the  denominator  the 
result  will  be  J- 1 

Let  x  denote  the  numerator,  and  y  the  denominator. 

From  the  conditions  of  the  problem, 
x+  1       1 


x  1 


or         Sx  -f-  3  =  y, 
or         4x  =  1  +  y  ; 


l  +y     4 

hence,  by  combination,    x  =.  4     and     y  =  15.     Ans.  ^-. 

21.  A  shepherd  was  plundered  by  three  parties  of  soldiers.  The 
first  party  took  -J-  of  his  flock  and  i  of  a  sheep  ;  the  second  took  £ 
of  what  remained  and  J  of  a  sheep ;  the  third  took  £  of  what  then 
remained  and  J  of  a  sheep,  which  left  him  but  25  sheep :  how  many 
had  he  at  first  1 

Let  x  denote  the  number  of  sheep.  Then,  after  being  plundered 
by  the  1st  party,  he  would  have 


-(l  +  i)  =  — 4—   s^Pi 


after  being  plundered  by  the  2d  party,  he  would  have 
Sx  —  1        Ar  —  1    .    1\       x  —  1 


4 


V     12     r  51 


166  APPENDIX. 

after  being  plundered  by  the  3d  party,  he  would  have 
x  —  I       (x  —  1       1\       ar— 3 


-p?+i> 


2 
from  the  conditions  of  the  problem, 

x  —  3 
■         =  25,     or     x  —  3  =  100 ;         :\     *  =  103. 

22.  What  two  numbers  are  those  whose  product  is  63,  and  the 
square  of  whose  sum  is  equal  to  64  times  the  square  of  their  dif 
ference  ? 

Let  x  and  y  denote  the  two  numbers. 
From  the  conditions  of  the  problem, 

*y  =  63 (i), 

(*  +  y)»  =  64  («  -  y)»     •     ,     (2); 

extracting  the  square  root  of  both  members  of  (2), 

x  +  y  =  8(x  —  y),     or     7*  =  9y;         .-.     x  =  $ y  ; 
substituting  in  (1),  %y2  =  63 ; 

.  • .     y2  =  49     and     y  =  7,     also     x  =  9. 

23.  The  sum  of  two  numbers  multiplied  by  the  greater  gives 
209 ;  their  sum  multiplied  by  their  difference  gives  57 :  what  are 
the  two  numbers  1 

Let  x  and  y  denote  the  numbers. 
From  the  conditions  of  the  problem, 

{x  +  y)  x  =  209,     or     x2  +  xy=  209     .     .     (1), 
(x  +  y)(x-y)^    57,     or     x2  -  y2  =    57     .     .     (2); 
subtracting  (2)  from  (1),  member  from  member, 
xy  +  y2=  152     .     .     (3) ; 
adding  (3)  and  (1),  member  to  member, 

x2  +  2*y  +  y2  =  361 ;         .-.     x  +  y  =  19-, 


ADDITIONAL  EXAMPLES.  167 

209 
hence,  from  (1),  x  =  — -  =  11 ;     also,     y  =  8. 

iy 

24.  Three  numbers  are  in  arithmetical  progression  ;  their  sum  is 
15,  and  the  sum  of  their  cubes  is  495  :  what  are  the  numbers  % 
Let  ar,  y  and  z  denote  the  numbers, 
From  the  conditions  of  the  problem, 

y   —  x   =  z    —      y     .     .     (1) 
*  +y  +z  =    15     .     .     (2) 
x3  +  y3  +  z3  =  495     .     .     (3) ; 
from  (1),         2y  =  z  -f  x,       which  in  (2),  gives       y  =  5 ; 
substituting  in  (2)  and  (3), 

z   +   x  =    10     .     .     (4) 
z3  +  x3  =  370     .     .     (5)  ; 
dividing  (5)  by  (4),  member  by  member, 

Z2  __  zx  +  X2  £J  37       ,       c        ^  . 

squaring  both  members  of  (4), 

Z2  +  2zx  +  x*  =  100     .     .     (7)  ; 
combining  (6)  and  (7), 

3^  =  63,     or     zx  =  21:         .'.      z  =  — ; 
substituting  in  (4), 

re  +  —  =  10,       or      z2  -  10a;  =  -  21 : 


.-.     x  ==  5  db  </—  21  -f  25  =  5  ±  2  ;         hence,     x  =  7,     or     3. 

25.  Divide  the  number  16  into  two  parts  such  that  25  times  the 
square  of  the  first  shall  be  equal  to  9  times  the  square  of  the 
second. 


168  APPENDIX. 

Let  *  denote  one  part ;  then  will    16  —  *   denote  the  other. 
From  the  conditions, 

25z2  =  9  (256  -  32*  +  *2)  =  2304  -  288*  +  9*2 ; 

reducing,  x2  -f-  18*  =  144  ; 

.-.     *  =  —  9  ±  -/144  +  81  =  —  9  ±  15,     or     *  =  6, 

since  the  negative  value  does  not  satisfy  the  problem  understood  in 
the  numerical  sense. 

26.  There  are  two  numbers  such  that  the  greater  multiplied  by 
the  square  root  of  the  less  is  18,  and  the  less  multiplied  by  the 
square  root  of  the  greater  is  12  :  what  are  the  numbers  I 

Let  *  and  y  denote  the  numbers. 

From  the  conditions  of  the  problem, 

y^x~=\8     .     .     (1) 

*V7=12     .     .     (2); 

i 

multiplying  (1)  by  (2),  member  by  member, 


V(*^=216;         .'.     *y  =  36     .     .     (3); 
adding  (1)  and  (2),  member  to  member, 

or  V*  +  vV  =  5 ; 

squaring  both  members, 

*-fy  + 2-^=25     •     •     (5)> 
or  ff  +  JfsU    .    .     (6); 

oombining  (3)  and  (6), 

*  =  9.        y  =  4. 


ADDITIONAL   EXAMPLES.  169 

27.  What  two  numbers  are  those  the  square  of  the  greater  of 
which  being  multiplied  by  the  lesser  gives  147,  and  the  square 
of  the  lesser  being  multiplied  by  the  greater  gives  63  1 

Let  x  and  y  denote  the  numbers. 
From  the  conditions  of  the  problem, 

x2y  =  147     .     .     (1) 
xy*  =    63     .     .     (2); 
multiplying  (1)  and  (2),  member  by  member, 

x3y*  =  9261     .     .     (3), 
or  xy  =      21     .     .     (4) ; 

dividing  (2)  by  (4),  member  by  member, 

y  =  3  ;         in  like  manner,         x  =  7. 

This  method  of  solution  might  be  applied  to  the  equations  of  the 
preceding  example. 

28.  There  are  two  numbers  whose  difference  is  2,  and  the  product 
of  their  cubes  is  42875  :  what  are  the  numbers  ] 

Let  x  and  y  denote  the  numbers. 

From  the  conditions  of  the  problem, 

x-y  =  2     .     .     (1) 
*V  =  42875     .     .     (2); 
extracting  the  cube  root  of  both  members  of  (2), 

35 

xy  =  So;         .'.     y  =  —'y 
x 

substituting  and  reducing, 

x*  —  2x  =  35, 

x   =  1  ±  -/35  +  1  =  1  ±  6 ; 
.  • .     x  =  7,     and     —  5,     y  =  5,     and     —  7. 


170  APPENDIX. 

29..  A  sets  out  from  C  towards  D,  and  travels  8  miles  each  day ; 
after  he  had  gone  27  miles,  B  sets  out  from  D  towards  C,  and  goes 
each  day  ^  of  the  whole  distance  from  D  to  C;  after  he  had 
travelled  as  many  days  as  he  goes  miles  in  each  day,  he  met  A  : 
what  is  the  distance  from  D  to  C? 

Let  x  denote  the  number  of  miles  from  D  to  C. 

Then,  —  will  denote  the  number  of  miles  B  travels  per  day, 
also  the  number  of  days  that  he  travels  ; 


x2 
hence,     — —       denotes  the  number  of  miles  travelled  by  B, 

27  +  Sx       "         "         "  "  "  A. 

From  the  conditions  of  the  problem, 
x2  Sx 

clearing  of  fractions  and  reducing, 

x2  —  240z  ta  —  10800 ; 
.-.     x  -  120  ±  -/—  10800  +  14400  =  120  ±  60 ; 
whence,  x  =  60,         x  =  180. 

30.  There  are  three  numbers ;  the  difference  of  the  differences  of 
the  1st  and  2d,  and  2d  and  3d,  is  4 ;  their  sum  is  40,  and  their  con 
tinned  product  is  1764  :  what  are  the  numbers  1 

Let  a?,  y  and  z  denote  the  numbers. 

From  the  conditions  of  the  problem, 

(#-y)-(y-*)=      4    .    .    (l) 

x  +  y  +  z=      40     .     .     (2) 
xyz  =  l7U     .     .     (3); 


ADDITIONAL   EXAMPLES.  171 

combining  (1)  and  (2),  eliminating  x  and  s, 

3y  =  36;         .-.     y  =  12  ; 
substituting  in  (2)  and  (3), 

x  +  z  =    28     .     .     (4) 
xz  =  147     .     .      (5) ; 
combining  (4)  and  (5), 

*  ss  7,     or     21  ;         y  =  21,     or    7. 

31.  There  are  three  numbers  in  arithmetical  progression  :  the 
sum  of  their  squares  is  93,  and  if  the  first  be  multiplied  by  3,  the 
second  by  4,  and  the  third  by  5,  the  sum  of  the  products  will  be  66 : 
what  are  the  numbers  ? 

Let  x  denote  the  first  number,  and  y  their  common  difference. 

From  the  conditions  of  the  problem, 

x*  +  (x  +  yY  +  (x  +  2yy  =  9Z  .  .  (1) 

3a  +  4(s  +  y)  +  5(*  +  2y)  =  66  .  .  (2); 
performing  indicated  operations  and  reducing, 

3*2  4-  5y2  +  6*y  =  93  .  .  (3) 

12*  +  14y  =  66,     or     6*  -f-  7y  =  33  .  .  (4). 

■^                                               33  — 6a? 
From  (4),  y  = = ; 

1089  —  396*  +  36z2              a                  33*  —  6*2 
.-.     y2  = _ ,  and         xy  = _ . 

substituting  in  (3)  and  reducing, 

2       198  296 

*   -^5*-~~  25; 


172  APPENDIX. 

99  /     296~      9801       99  ±  49 

whence,  *  =  -  ±W  -  —  +  —  = 


25       ' 

148 

.*.     x  = ——-,         x  =  2. 
25 

Taking  the  second  value  of  x,  we  find    y  =  3,     and  the  numbers 
are  2,  5  and  8. 

The  problem  supposes  the  numbers  entire,  therefore  the  1st  value 
of  x  is  not  used. 

32.  There  are  three  numbers  in  arithmetical  progression  whose 
sum  is  9,  and  the  sum  of  their  fourth  powers  is  353  ;  what  are  the 
numbers  ] 

Let  x,  y  and  z  denote  the  numbers. 
From  the  conditions  of  the  problem, 

2y  =  x  +  z     .     .     (1) 
x  +  y  +  z  =  9     .     .     (2) 
a;Hy4  +  24  =  353     .     .     (3). 
From  (1)  and  (2)  we  find  y  =  3  ; 

substituting  in  (2)  and  (3), 

x   +  z   =      6     .     .     (4) 
a*  +  z4  =  272     .     .     (5)  ; 
raising  both  members  of  (4)  to  the  4th  power, 

X4  _|_  4x3z  _j_  qx2z2  +  4XZ3  +  2*  _  1296     .     .     (6) ; 

adding  equations  (5)  and  (6),  member  to  member,  and  dividing  by  2 

x*  4  2x3z  4  3z222  +  2z23  4  -4  =  784     .     .     (7); 
extracting  the  square  root  of  both  members, 

x*  4  xz  4-  s2  =  28     .     .     (8)  ; 


ADDITIONAL   EXAMPLES.  173 

squaring  both  members  of  (4), 

x2  +  2xz  -f  z2  =a  36     .     .     (9); 

from  (8)  and  (9)  we  find 

M&8     .     »     (10); 

from  (4)  and  (10)  we  get 

#  =s=  2,     or    4 ;  «  =  4,     or    2: 

hence,  the  numbers  are  2,  3  and  4. 

33.  How  many  terms  of  the  arithmetical  progression  1,  3,  5,  7, 
&c,  must  be  added  together  to  produce  the  6th  power  of  12? 
The  Gth  power  of  12  is  2985984. 
From  Art.  176  we  have  the  formula, 


d~2a  dfc  y/(d-Za)2  +  §dS 
n  m g 

Ir.  the  present  case,         o=l,     d  —  2,     and     8  aa  2985984  ; 


i     •      •  -v/1*5  X  2985984       IMi 

substituting,         n  ss  — — 2*  1728. 

34.  The  sum  of  6  numbers  in  arithmetical  progression  is  48  ;  the 
product  of  the  common  difference  by  the  least  term  is  equal  to  the 
number  of  terms  :  what  are  the  terms  of  the  progression  1 

Let  x  denote  the  1st  term,  and  y  the  common  difference. 

"From  the  conditions  of  the  problem, 

c 

Gx  -f  15y  ss  48,     xy  zz  6  ;         .    .     y  -=  - ; 

x 

substituting  and  reducing, 

x2  —  8x  ss  -»  15; 


.-.     X  ae  4  dfc  V-  15  -f-  16  =  4  dfc  1,     or    a;  =  5,     a;  =  3 } 
whence,  y  =  £ ,         y  =  2 ; 


174  APPENDIX. 

hence,  the  series  is       3.5.7.9.11.13, 
or  5.6±.7£  .  8f  .  9f  .11. 

35.  What  is  the  sum  of  10  square  numbers  whose  square  roots 
are  in  arithmetical  progression  the  least  term  of  which  is  3,  and  the 
common  difference  2  ? 

Let  x  denote  the  sum. 

The  progression  of  roots  is 

3.5.7.9.11.  13.  15.17.19.21, 
and  the  series  of  squares, 

9  .  25  .  49  .  81  .  121  .  169  .  225  .  289  .  361  .  441. 
1st  order  of  diffs,  16,  24,  32,  40,  &c, 

2d  order  of  diffs,  8,  8,  8,  &c, 

3d  order  of  diffs,  0,  0,  &c. 

Frgm  Art.  210,  making  ' 

S'  =  x,     a  =  9,     7i  =  10,     c?!  =  16,     d2  —  S,     d3  =  0,     <&c. 
x  ==  90  -f  45  x  16  +  120  x  8  --=  1770. 

36.  Three  numbers  are  in  geometrical  progression  whose  sum  is 
95,  and  the  sum  of  their  squares  is  3325 :  what  are  the  numbers? 

Let  #,  y  and  z  denote  the  numbers. 

From  the  conditions  of  the  problem, 

y2  =  xz     .     .  (1) 

x2  +  y2  +  *2  =  3325     .     .  (2) 

x  +y   +z   =      95     .     .  (3); 
combining  (1)  and  (2), 

x*  +  2xz  +  z2  =  3325  +  m  .     .     (4)  ; 


ADDITIONAL   EXAMPLES.  175 

combining  (1)  and  (3), 

x  +  y'xz  +  z  =  95     .     .     (5) 
from  (4)  and  (5), 

x  4-  t  =  -y/3325  +  a* 

a;  4-  z  =  95  —  ^$bb ; 

hence,  -y/3325  -f  #2  =  95  —  y»T; 

squaring  both  members, 

3325  +  xz  —  9025  —  190  ^/xz  +  *z ; 

.  • .     y/x~z  =  30,     or     xz  -.  900     .     .     (6)  ; 

substituting  in  (5),  a;  +  z  as    65     .     .     (7)  • 

from  (6)  and  (7),         *  =  20     and    45, 

y  =  45     and     20. 

37.  Three  numbers  are  in  geometrical  progression  :  the  difference 
of  the  first  and  second  is  6 ;  that  of  the  second  and  third  is  15 : 
what  are  the  numbers  1 

'Let  x.  y  and  z  denote  the  numbers. 

From  the  conditions  of  the  problem, 

y2  =  xz     .     .      (1) 
x  —  y  =  —    6 ;         .".     x  =  y  —    6 
y  —  z=i  —  15;         .'.     2=^4-15, 
and  xz  =  y2  -f-  9y  —  90; 

substituting  in  (1)  we  find         y  =  10  ; 

.  • .     x  =z  4,     and     2  =  25. 

38.  There  are  three  numbers  in  geometrical  progression  ;  the  sum 


176  APPENDIX. 

of  the  first  and  second  is  14,  and  the  difference  of  the  second  and 
third  is  15  :  what  are  the  numbers? 
Let  #,  y  and  z  denote  the  numbers. 
From  the  conditions  of  the  problem, 
y2  as  xz     .     .     (1) 
#  +  y  =  14;         .*.     x  =e  14  —  y 
z  —y  =  15;         .-.     z  =  15  +  y, 
and 

substituting  in  (1), 


g2  t=  210  —  y  —  y2 ; 
^  +  |  =  105; 

and 

'.»5+ ,'„..-':«=.., 

21 
~  2"} 

taking  the  1st  value  of  y,  we  find 

x  3s  4i         *  =  25. 

39.  A,  B  and  C  purchase  coffee,  sugar  and  tea  at  the  same  prices; 
A  pays  $11,621  for  71  pounds  of  coffee,  3  pounds  of  sugar,  and  2} 
pounds  of  tea;  B  pays  $16,25  for  9  pounds  of  coffee,  7  pounds  of 
sugar,  and  3  pounds  of  tea;  C  pays  $12,25  for  2  pounds  of  coffee, 
5J  pounds  of  sugar,  and  4  pounds  of  tea :  what  is  the  price  of  a 
pound  of  each  1 

Let  x,  y  and  z  denote  the  number  of  cents  that  the  coffee,  sugar 
and  tea  cost,  respectively. 

^rom  the  conditions  of  the  problem, 

7J&+    3y  +  2l2  =  1162^  *  •  0) 

9*+    7y  +  3z=  1625  •  •  (2) 

2  z  +  5Jy  +  4  2  =  1225  •  .  (3) ; 


ADDITIONAL  EXAMPLES.  177 

clearing  (1)  and  (3)  of  fractions, 

SOx  f  12y  +  9*    =4650      •     •    (4) 
4  x  +  lly  +  8z    =2450       •     •     (5). 

From  (2)  and  (4), 

Zx  —  9y  =  —  225,     or    x  —  Sy  =  —  75     •     •     (6) ; 

from  (2)  and  (5), 

60s  +  23y  =  5650     .     -     (7); 

from  (6)  and  (7),  y  =  50; 

by  substitution,  s  =  75,  z  =  200. 

40.  Divide  100  into  2  such  parts  that  the  sum  of  their  square 
roots  shall  be  14. 

Let  x  denote  the  first  part. 

From  the  conditions  of  the  problem, 


<y/x  +  -/100  —  a;  =  14; 
squaring  both  members  and  reducing, 

v/100s-z2  =  48; 
squaring  both  members  and  reducing, 

z2-  100s  =  -2304; 


.\     *  =  50  ±  -/-  2304  +  2500  =  50  ±  14, 
a;  =  64,    and    36. 

41.  In  a  certain  company  there  were  three  times  as  many  gentle- 
men as  ladies ;  but  afterwards  8  gentlemen  with  their  wives  went 
away,  and  there  then  remained  five  times  as  many  gentlemen  as 
ladies :    how  many  gentlemen,  and  how  many  ladies  were  there 

originally  ? 

12 


178  APPENDIX. 

Let  Sx  denote  the  number  of  gentlemen  ;  then  will  x  denote  the 
number  of  ladies. 

From  the  conditions  of  the  problem, 

Sx  -  8  =  5  {x  -  8) ; 

.-.     x  =  16,     and     3#  =  48. 

42.  Find  two  quantities  such  that  their  sum,  product,  and  the 
difference  of  their  squares,  shall  all  be  equal  to  each  other. 

Let  x  and  y  denote  the  quantities. 
From  the  conditions  of  the  problem. 

x   +  y   =  xy     •     •     (1) 
x2  —  y2  =  xy     •     •     (2)  ; 
by  division  of  (2)  by  (1),  we  have 

x  —  y  =  1,     or     x  =  y  +  I-, 
substituting  in  (1), 

2y  +  1  =  y2  +  y,     or     y2  —  y  =  1  ; 

whence,  y  =  2      V         ?     °r    V  ~ 2 ; 

hence,  x  ==  — rr— 

43.  A  bought  120  pounds  of  pepper,  and  as  many  pounds  of 
ginger,  and  had  one  pound  of  ginger  more  for  a  dollar  than  of 
pepper ;  the  whole  price  of  the  pepper  exceeded  that  of  the  ginger 
by  6  dollars :  how  many  pounds  of  pepper,  and  how  many  of 
ginger  had  he  for  a  dollar1? 

Let  x  denote  the  number  of  pounds  of  pepper  for  a  dollar. 


ADDITIONAL   EXAMPLES*  179 


From  the  conditions  of  the  problem, 


120  120  P  2^  OA 

=  6,     or     x2  -f  x  =  20 ; 

x         x  -f-  1 


.  • .     «r=—  -db\/20  4--  =  — * — :     hence,     x  =  4. 
2       V  4  2       ' 

The  negative  value  does  not  conform  to  the  conditions  of  the 
special  problem. 

44.  Divide  the  number  36  into  3  such  parts  that  the  second  shall 
exceed  the  first  by  4,  and  that  the  sum  of  their  squares  shall  be 
equal  to  4G4. 

Let  x,  y  and  z  denote  the  parts. 
From  the  conditions  of  the  problem, 

x  +  y  +  z-SQ     .     .     (1) 
y-x  =  4       •     .     (2) 
xl  +  y2  +  z2  =  464  •     ♦     (3) ; 
from  (1),  x2  +  2xy  +  y2  =  1296  -  T2z  +  z*     .     .     (4)  ; 

from  (2),  x2  -  2xy  +  y2  -  16        (5)  ; 

adding  (4)  and  (5),  member  to  member, 

2x2  +  2y2  =  1312  —  72s  -f  z2     i     .     (6)  ; 
from  (3),  2x2  +  2y2  =    928  -    2z2  .     .     .     .     (7) ; 

equating  the  second  members  and  reducing, 
z2  -  24s  se  -  128  : 


,\     *=12±  ^128+  144  =  12dfc4; 

hence,  e  =  16,         s  =  8 ; 

substituting  the  first  value  in  (1), 

*  +  y-20     .     .     (8); 


180  APPENDIX. 

from  (2)  and  (8),  y  =  12    and    x  =  8. 

45.  A  gentleman  divided  a  sum  of  money  among  4  persons,  so 
that  what  the  first  received  was  J  that  received  by  the  other  three , 
what  the  second  received  was  ^  that  received  by  the  other  three ; 
what  the  third  received  was  £  that  received  by  the  other  three,  and 
it  was  found  that  the  share  of  the  first  exceeded  that  of  the  last  by 
$14  :  what  did  each  receive,  and  what  was  the  whole  sum  divided  % 
Let  x,  y,  z  and  w  denote  the  number  of  dollars  that  each  received. 
From  the  conditions  of  the  problem, 

2x  =  y  +  z  +  w  •  <  (1) 
Sy  =s  x  -f  z  -f  w  •  *  (2) 
42  =  x  -f-  y  +  io    «     *     (3) 

x  —  w  =  14     i     •     (4)  j 
from  (2)  and  (3), 

x  -\-  w  z=  Sy  —  2 

x  -f  w  =  4sr  —  y  ;     whence,     3y  —  2  =.  42  —  y, 
oi  4y  =  52,         *  =  |y     •     •     (5); 

from  (4),  w  =  x  —  14     •     •     (6) ; 

substituting  the  values  of  w  and  z  in  (1)  and  (2), 

2ar  =  y  -f-  f  y  +  x  —  14 

3y  =  a;-f-|y-f^  —  14;     whence,  by  reduction, 
5x  —    9y  ae  —  70 

,  \     a;  =  40,     y  as  30 ;     and  by  substitution,    2  =  24,     w  =  20. 

46.  A  woman  bought  a  certain  number  of  eggs  at  2  for  a  penny, 
and  as  many  more  at  3  for  a  penny,  but  on  selling  them  at  the  rate 


ADDITIONAL   EXAMPLES.  181 

of  5  for  2  pence,  she  lost  4  pence  by  the  bargain ;  how  many  did 
she  buy  ?     . ,. 

Let  x  denote  the  number  at  each  price.     Then  will     -  -f  :  — 

2(x  4-  x) 
denote  the  number  of  pence  paid,  and     — — » — =      will  denote  the 

o 
number  of  pence  received. 

From  the  conditions  of  the  problem, 

x       x       %(x  4-  .r) 

2  +  3  =  ~5 +  4  '         reducing>         *  =  12°- 

47.  Two  travellers  set  out  together  and  travel  in  the  same  direc- 
tion ;  the  first  goes  28  miles  the  first  day,  26  the  second  day,  24  the 
third  day,  and  so  on,  travelling  2  miles  less  each  day ;  the  second 
travels  uniformly  at  the  rate  of  20  miles  a  day  :  in  how  many  days 
will  they  be  together  again  ] 

Let  x  denote  the  required  number  of  days.  The  distance 
travelled  by  the  first  in  x  days  is 

[(Art.  176),  since     a  =  28,     d  =  —  2,     and     n  =  #],     denoted  by 

£s[56  —  (s  -  I)2],     or     29*  -z2; 

and  the  distance  travelled  by  the  second  is  denoted  by  20a?: 
hence,  we  have 

29x  —  x'1  =z  20*,     or     x  =  9. 

48.  A  farmer  sold  to  one  man  30  bushels  of  wheat  and  40  of 
barley,  for  which  he  received  270  shillings.  To  a  second  man  he 
sold  50  bushels  of  wheat  and  30  of  barley,  at  the  same  prices,  and 
received  for  them  340  shillings  :  what  was  the  price  of  each  1 

Let  x  denote  the  number  of  shillings  for  1  bushel  of  wheat, 
and      y       "         "         "  "  "        "  barley. 


182  APPENDIX. 

Prom  the  conditions  of  the  problem, 

30*  +  40y  =  270     •  .     (1) 

50x  +  30y  =  340     •  •     (2) ; 
whence,                 HOy  =  330,     or    y  =  S;      hence,     x  =  5. 

49.  There  are  two  numbers  whose  difference  is  15,  and  half  their 
product  is  equal  to  the  cube  of  the  lesser  number  :  what  are  the 
numbers  1 


Let  x  and 

y  denote  the  numbers  ; 

from  the  conditions  of  the  problem, 

x  —  y  = 

15 

xy  =  2y3 ;             or, 

substituting 

and  reducing, 

v 

2       1 

15 
2  ' 

*  =  2y2; 


1  /15        1         1±  11 

•'•     y  =  4±V^  +  l6  =  ""4— ; 

5 

hence,        y  =  3,     and     —  - ;     also,     x  =  18,     and 


25 


50.  A  merchant  has  two  barrels  and  a  certain  number  of  gallons 
of  wine  in  each.  In  order  to  have  an  equal  quantity  in  each,  he 
drew  as  much  out  of  the  first  cask  into  the  second  as  it  already- 
contained  ;  then  again  he  drew  as  much  out  of  the  second  into  the 
first  as  it  then  contained  :  and  lastly,  he  drew  again  as  much  from 
the  first  into  the  second  as  it  then  contained,  when  he  found  that 
there  was  16  gallons  in  each  cask:  how  many  gallons  did  each 
originally  contain  1 


ADDITIONAL   EXAMPLES.  183 

Let  x  denote  the  lumber  of  gallons  in  the  first  cask,  and  y  the 
number  in  the  second  ; 

x  —  y  will  denote  the  quantity  in  the  first  cask  after  the  first  drawing, 
and  2y  the  quantity  in  the  second  cask ;  after  the  second  drawing, 
2y  —  (x  —  y)  or  Sy  —  x  will  denote  the  quantity  in  the  second, 
and  2x  —  2y  the  quantity  in  the  first  cask ;  after  the  third  drawing, 
2x  —  2y  —  (Sy  —  x)  or  Sx  —  5y  will  denote  the  quantity  in  the 
first  cask,  and  6y  —  2x  the  quantity  in  the  second. 

From  the  conditions  of  the  problem, 

Sx  -  6y  =  16 
6y  -  2x  =  16. 
By  combination, 

x  =  22;  y  =  10. 


184 


APPENDIX. 


PROBLEMS  FROM  LEGENDRE. 

Problem  VI. — In  a  right  angled  triangle,  having  given  the  base 
and  difference  between  the  hypothenuse  and  perpendicular,  to  find 
the  sides. 

Let  a  =  the  base  AB, 

b  =  the  difference  between  AC  and  BC, 
•  x  =  the  hypothenuse  AC,  and 
y  =  the  perpendicular  CB. 
Then,   x  —  y  =  b,  by  the  conditions,     .... 
z2  =  a2  +  y2(Bk.  IV.  Prop.  11),    .     .     . 
From  equation  (1)  we  have, 

x  =  y4-5; 
by  squaring,  x2  =  y2  -f-  2by  -f  b2. 

Substituting  this  value  of  x2  in  equation  (2), 

y2  +  2by  +  b2  =  a2  +  y2 ; 

a2  -b2 


hence,         2by  =  a54  —  b2,        or, 


2/ 


26    ; 

substituting  this  value  in  equation  (1),  we  readily  find 

a2  +  b2 

x  =  — — 

26 


Problem  VII. — In  a  right  angled  triangle,  having  given  the 
hypothenuse  and  the  difference  between  the  base  and  perpendicular,  to 
determine  the  triangle. 

Let  a  =  the  hypothenuse  AC, 

b  =  difference  between  AB  and  BC, 


PROBLEMS   FROM   LEGENDRE. 

1* 

35 

x  =  the  base  AB,         and 

C 

y  =  the  perpendicular  BC. 

/ 

Then,    x  —  y  =  ft,  by  the  conditions,       •     •     (1) 

/ 

a2  L  x2  +  y\  (Bk.  IV,  Prop.  11)      .     (2) 

A                   JB 

From  equation  (1),  we  have, 

x  —  y  +  ft; 

and  by  squaring, 

x2  =  y2  +  26y  +  62, 

Substituting  this  value  of  x2  in  equation  (2),  we  have, 

a2  =  y2  +  2by  +  b2  +  y2-, 
and  by  transposing  and  reducing, 

....  «2  -  b2 

V2  +  by  =  — - — ; 

-  ft  +  \/2a2  -  ft2                         /ft  4-  t/2a2  -  ft2\ 
hence,  y  = =g ;         y  =  -  ^ v-- J  ; 

Substituting  either  of  these  values  in  equation  (1),  we  readily  find 
the  corresponding  value  of  x. 

Note. — The  positive  value  of  the  unknown  quantity  generally  fulfils  the  con- 
ditions of  the  problem,  understood  in  its  arithmetical  sense. 

The  negative  value  will  always  satisfy  the  conditions  of  the  equation:  with 
its  sign  changed,  it  may  be  regarded  as  the  answer  to  a  problem  which  differs 
from  the  one  proposed  only  in  this :  that  certain  quantities  which  were  additive 
have  become  subtractive,  and  the  reverse. 

Problem  VIII. — Having  given  the  area  of  a  rectangle  inscribed  in 
a  given  triangle,  to  determine  the  sides  of  the  rectangle. 

By  a  given  triangle,  we  mean  one  whose  sides  are  all  known,  or 
given. 

1st.   To  find  the  perpendicular  and  segments  of  the  base: 


186 


APPENDIX. 


Let  ABC  be  a  triangle,  in  which  the  three 
sides  are  given  :  viz., 

b  =  AB,     a  =  AC,     and     c  =  BC. 
Let  CD  be  drawn  perpendicular  to  AB,  and      £ 

let       y  =  CD,     and     x  z=  AD  ;     then  will     b  —  x  =  DB. 

r 

Then,  a2  ==  x2  -f-  y2 (1) 

c2  =y2  +  x2  —  2bx  +  b2     .     .     •     (2)  ; 

substituting    in   equation   (2)   the  value    of    x2  +  y2  =  a2,     from 
equation  (1),  and  transposing, 

c2  =  a2  —  2bx  -f  52 ;     hence, 
a2  +  5>  _  c2 


a:  = 


26 


and  substituting  this  value  in  equation  (1),  we  find  the  altitude  of 
the  triangle. 

2d.   To  find  the  sides  of  the  rectangle. 
Suppose  the  rectangle  to  be  inscribed  in  the  c 

triangle  ACB. 

Let  d  =  the  area  of  the  rectangle, 

x  =  its  base,  * 

y  =  its  altitude,  and  A 

h  =  CD,  the  altitude  of  the  triangle. 


Then,  by  similar  tr 
AB 
b 

and 


angles, 

CD     :  :     GF     :     CJ;     that  is, 
h      :  :       x       :     h  —  y\  hence, 
hx  =  bh-by     •     •     •     (1), 
xy  =  c?,     by  the  conditions,  (2). 
d 


From  equation  (2),  we  have         x  =  - ; 


PROBLEMS  FROM  LEGENDRE. 


187 


substituting  this  value  in  equation  (1),  we  have, 


h-  =  bh  —  by;     clearing  of  fractions, 

h  d  ~  bhy  —  by2  ;     or, 
-hd 


hy 


»=+2  + 


w- 


b  h2  -  4M 


whence, 
J1 


b  h2  -  4hd 


b  '   2      2V  b 

Substituting  these  values  of  y  in  equation  (1),  we  find  the  corres- 
ponding values  x. 

Problem  IX. — In  a  triangle,  having  given  the  ratio  of  the  two  sides, 
together  with  both  the  segments  of  the  base  made  by  a  perpendicular 
from  the  vertical  angle,  to  determine  the  triangle. 

Let  ACB  be  a  triangle,  and  CD  a  line  drawn  perpendicular  to 
the  base  AB. 


Let 


Then, 


AC  =  x 
CB=  y 
CD=  i 
AD=  a 
DB=  b 
c—  ratio. 

-  =  c,     by  the  conditions 

y 


x2  —  a?  =  z2  (Bk.  IV.,  Prop.  11,  Cor.  1)     .     . 

y2 —  b2  =  z2(Bk.  IV.,  Prop.  11,  Cor.  1)     .     . 

Subtracting  (3)  from  (2),  member  from  member, 


(i) 

(2) 
(3). 


y2  +  J2  —  a2  =  0  ;     cr,     x2  =  y2  +  a2  —  i2 


(4). 


188  APPENDIX. 

From  equation  (1)  we  have 

x  =  cy ;         hence,         x2  =  c2  y2       •     •     •     (5)  j 
combining  (4)  and  (5),  we  have 

C2y2  _  yl  _j_  a2  _   ^       and       (C2  _   1)  y2  _  a2  _  J2  . 

'»  a2-^2  ,  /a2  -  62 

hence,  y2  =  — — p  ;      and     y  =  ±\J ^—-y ; 

substituting  these  values  in  equation  (1),  we  find  the  corresponding 
values  of  x. 

Problem  X. — In  a  triangle,  having  given  the  base,  the  sum  of  the 
other  two  sides  and  the  length  of  a  line  drawn  from  the  vertical  angle 
to  the  middle  of  the  base,  to  find  the  sides  of  the  triangle. 

Let  ABC  be  a  triangle,  and  CD  the  line  drawn  from  the  vertex 
C  to  D,  the  middle  point  of  the  base. 

Let  AC  =  x  \ 

BC  =y 

AD  =  b 

CD  -a 

AC  -f  BC  =V 

Then,  x  -f  y  =  s,     by  the  conditions     •     •     (1) 

and         x2  -f  y2  =  262  +  2a2,  (Bk.  IV.,  Prop.  14)     •     •     (2) ; 

equation  (1),  by  transposing  and  squaring,  gives 

y2  =  x2  —  2sx  +  s2 ; 

substituting  this  value  in  equation  (2),  we  have 

2x2  —  2sx  +  s2  =  2b2  +  2a2 ; 


PROBLEMS  FROM  LEGENDRE,  189 


transposing  and  reducing,  -we  have 


262  -f-  2a3  -  s2 
s?  —  sx  zz ;     whence, 


s  +  V*t>2  +  4a2  —  *2           %              ^— -v/462-f4a2— «2 
x  = - ,     and     x  as 


Problem  XI.— In  a  triangle,  having  given  the  two  sides  about  the 
vertical  angle,  together  with  a  line  bisecting  that  angle  and  terminating 
in  the  base,  to  find  the  base. 

Let  ACB  be  a  triangle,  and  CD  a  line  bisecting  the  angle  ACB, 
and  terminating  in  the  base  at  D. 


Let 

AC   =za 

C 

BC  =  6 

A\ 

CD  =  c 

/W 

AT>  =  x 
DB  =  y. 

I  \  \ 

A        i)            1 

Then,        x 

!   y 

:  :     a     :     b 

(Bk. 

IV., 

Prop.  17)  ; 

whence, 

bx 

23  ay 

'  •  •  •   (i); 

also,         a  x  b  -  c2  +  *y  (Bk.  IV.,  Prop.  31)     •     •      (2). 

Multiplying  the  first  equation  by  x,  and  the  second  by  a,  and  sub- 
tracting, member  from  member,  we  have 

bx2  —  a2b  =2  —  ac2  j  whence 


fa  (ah  _  C2)  A  (ab  -  c2 


Problem  XII. —  To  determine  a  right  angled  triangle,  having  given 
the  lengths  of  two  lines  drawn  from  the  acute  angles  to  the  middle 
points  of  the  opposite  sides. 


190  APPENDIX. 

Let  ABC  be  a  right  angled  triangle,  and  AD,  CE,  two  lines  drawn 
to  the  middle  points  D  and  E  of  the  opposite  sides. 
Let  AD  =  a 
CE  "=  b 
AB  =  2x 
BC  =  2y 
Then,  4x2  +  y2  =  a2  (Bk.  IV.  Prop.  11)       .     (1) 
and,     4y2  -f  x2  —  b2     "       "       "      "      .     .     (2). 
Multiplying  equation  (1)  by  4,  and  subtracting  (2)  from  it,  we  have, 


/AaT 
15z2  =  4a2  —  b2  ;         when,  x  =.-  dt  \I 

A 

and  substituting  in  equation  (2)        y  ts  sfc  */  - 


-&24 


462   -  rf 

15 


Problem  XIII. —  To  determine  a  right  angled  triangle,  having  given 
the  perimeter  and  the  radius  of  the  inscribed  circle. 

1st.  Let  ABC  be  a  right  angled  triangle,  O  the  centre,  and  r  the 
radius  of  the  inscribed  circle. 

Let  p  =  the  perimeter  ;  let  x  denote  the  length 
of  the  equal  tangents  drawn  from  A  (Bk.  III.  Prop. 
14,  sch.) ;  v  the  length  of  the  equal  tangents  drawn 
from  B ;  and  y  the  length  of  the  equal  tangents 
drawn  from  C. 

Then,  AC  =  x  -f-  y>     AB  sc #  -|-  v9     and  BC  =  y  +  v  ;     then, 
AC  +  AB  -f  BC  =  2x  +  2y  +  2v  =p ; 
transposing  and  reducing, 

x  +  y  =  ^—7; —  =  a>  a  known  quantity     •    •     •     (1). 


PR0BLEM8  FROM  LEGENDRE.  191 

Then,  AB2  +  to  =  AC2 :     that  is, 

x2  +  %vx  +  v2  -f  y2  +  2vy  +v2  =  a2  •  •  •  •  (2). 
2nd.  Observe  that  the  double  area  of  each  of  the  triangles  AOB, 
BOC,  and  AOC,  is  equal  to  its  base  multiplied  by  the  radius  of  the 
inscribed  circle ;  and  hence,  the  sum  of  these  products  is  equal  to 
the  sum  of  the  bases  multiplied  by  r ;  that  is,  =  r  x  p,  a  known 
quantity. 

But  the  base  AB  x  BC  is  also  equal  to  double  the  area  of  the 
triangle  ABC ;  hence, 

(x  +  v)  x  (y  -f  v)  =  r  x  p ;  that  is, 
xy  +  vx  +  vy  +  v2  —  r  X  p   .     .     .     .     (3)  ; 
Multiplying  both  numbers  of  equation  (3)  by  2,  we  have, 
2xy  -f-  2w  -h  2vy  -\-  2v2  =  2rp     .     .     .     (4)  ; 
subtracting  equation  (4)  from  (2),  we  have, 

x2  —  2xy  +  y2  =  a2  —  2rp ;  whence,  by  extracting  square  root, 

x  —  y  =  dfc  -y/a2  —  2rp  =  5 (5)  ; 

combining  (1 )  and  (5)  we  readily  find  the  values  of  x  and  y , 

a  +  b  ,  a  —  b 

*  =  - — ,     and     y  =  -2— 

Problem  XIV. —  To  determine  a  triangle,  having  given  the  base} 
the  perpendicular,  and  the  ratio  of  the  two  sides. 

Let  ACB  be  a  triangle,  and  CD  perpendicular  to  the  base  AB. 
Let  AC  =  y, 


then, 


r 

== 

ratio  ; 

ry 

as 

CB 

AB 

— 

b 

DB 

— 

b  -+X 

AD 

— 

X 

CD 

2S 

h. 

192 


APPENDIX. 


Then,         y2  =  x2  +  A2 ;     and     r2y2  a  b2  -  26*  -f  x2  +  A*. 
Multiplying  the  first  equation  by  r2,  and  subtracting, 

0  =  62  —  2bx  +  a;2  +  h2  —  r2*2  —  r2A2  ;     whence, 
(1  _  r2)  x2  -  2bx  a  (r2  -  1)  A2  -  62 ;     or, 


2b 


(r2-  1)  A2  —  62 
1  -r* 


whence, 


x  = 


b  a  y[(r2  -  1)  h2  -  b2]  (1  -  r2)  +  62 
1  -r2 


Problem  XV. —  To  determine  a  right-angled  triangle,  having  given 
the  hypothenuse,  and  the  side  of  the  inscribed  square. 

Let  ACB  be  a  right-angled  triangle,  and  FDEB  an  inscribed 
square. 

Let         AC  a  A,     and     DF  a  s :     also, 
denote  AB  by  x,  and  BC  by  y;  then  CE=y  — s. 
Then,         AB    :     BC    :  :     DE     :     EC ; 
that  is,  a;     :       y      :  :        5       :    y  —  5 ; 

whence,  xy  ~  sx  =  sy;     or, 

ary  =  sy  +  *r  a  .9(2-  +  y)     •     •     (1); 
also,  x2  +  y2  =  A2  (Bk.  IV.,  Prop.  11) 

If  to  equation  2,  we  add  twice  equation  (1),  we  have 
x2  +  2xy  +  y2  =  2s  (x  -f-  y)  +  A2 ;     or, 
(*  +  y)2-2s(z-by)=A2     .     .     (3), 
which  is  an  equation  of  the  second  degree,  in  which  the  unknown 
quantity  is  x  +  y  ;  hence, 

x  +  y  =  s  +  <y/h2  +  *2     •     •     (4). 
■ 
Combining  equations  (1)  and  (4),  we  have 

xy  a  s2  +  s  -/A2  -f  s2     •     •     (5). 


(2). 


PROBLEMS   FROM   LEGENDRE. 


193 


From  the  square  of  equation   (4),  subtract  4  times  equation  (5), 
and  we  have 


x2  —  2xy  +  y2  =  h2  —  2s2  —  2s  <y/h2  +  s2 
extracting  the  square  root  of  both  members, 


x  —  y  =\/h2  ~  2s2  —  2s  y^2  -f 
combining  equations  (4)  and  (7),  we  have 


(6); 
CO, 


x  e= 


y  = 


s  +  Vh2  +  *2  +  Vh2  -  2s2  —  2s  v/AMTs1 


s-f  -yA2  +  s2  —  ->A2  -  2s2  -  2s  -y/AM^P 


(8), 


(9). 


Problem  XVI. — To  determine  the  radii  of  three  equal  circles,  de 
scribed  within  and  tangent  to,  a  given  circle,  and  also  tangent  to  each 
other. 

Let  O  be  the  centre  of  the 
given  circle,  and  A,  B  and  C, 
the  centres  of  the  equal  in- 
scribed circles. 

Denote  the  radius  of  the 
given  circle  by  R,  and  the 
equal  radii  of  the  inscribed 
circles  by  r. 

Joining  the  centres  A  and 

C,  C  and  B,  B  and  A,  by 

Straight   lines,  we   have  the 

equilateral  triangle  ABC,  each 

of  whose  sides  is  2r.     Draw  COD  and  prolong  it  to  E,  and  it  will 

be  perpendicular  to  AB. 

13 


19±  APPENDIX. 

Then,  in  the  right-angled  triangle  ACD,  we  have 
CD2  =  AC2  -AD2,    or 
CD2  =  4r2  -  r2  =  3r2,     or     CD  =  r  y/S. 

But  since  0  is  the  point  at  which  lines  drawn  from  the  vertices  of 
the  angles  to  the  middle  points  of  the  opposite  sides,  in  both  tri- 
angles, intersect  each  other,  it  follows,  from  Cor.  of  Prob.  21,  that 

CO  =  f  CD  =  \r  V§";     hence, 

OE  =  R  =  r  +  Jr  y^ ;     hence,  finally, 

3R  R 

~3  +  2y^~  1  +2v4 

Problem  XVII. — In  a  right-angle  triangle,  having  given  the  peru 
'meter  and  the  perpendicular  let  fall  from  the  right  angle  on  the  hy- 
pothenuse,  to  determine  the  triangle. 

Let  ACB  be  a  right-angled  triangle,  right  angled  at  C ;  and  let 
CD  be  drawn  perpendicular  to  the  hypothenuse  AB. 

Let  p  =  the  perimeter,  and 

h  =  the  perpendicular  CD. 
Denote  A!C  by  ar,  and  CB  by  y  ; 
tken,  AB  =  -y/z2  -f  y2,     and  A 

AC-f-CB-f  ABrrrjt?;      that  is, 

z  +  y  +  tJx2  -f  y2=p    • 
xy  =  double  the  area  of  ACB     • 
h  -y/z2  +  y*~=  double  area     • 


Again, 

and 

hence, 


xy-h  V*2  +  y2 


(1). 

(2), 

(3); 

(4) 


PROBLEMS  FROM  LEGENDRE.  195 

transposing  in  equation   (1)  and  squaring, 

x2  -f  2xy  +  y2  =p*  —  2p  -/a2  -f  y2  +  x2  +  y2j 
or  2zy=^-%?V^rT7r    .     .     (5). 

multiplying  both  members  of  equation  (4)  by  2, 


2zy^2Ay*2  +  y2     .     .     .     (6); 
combining  equations  (5)  and  (6), 

2hy/x2  +  y«  =  ^2  _  2p^/X2  +  y2.       fo^ 

-         V^T7  =  2-^     •     .     .     (7); 


substituting  this  value  of  ^/x2  -f  y\  in  equation  (4), 

hP2 

squaring  both  members  of  equation  (7),  we  have 

x2  4-  v2  — P  ■       - 

adding  and  subtracting  2  times  each  member  of  (8), 

y     y  4(h+Py  TO 

Extracting  the  square  roots  of  equations  (9)  and  (10), 

r  |   V-P(P  +  M)    .  ,,„ 


y  8(A  +  j>)  I"'' 


196 


APPENDIX. 


hence, 


P  (P  +  2A)  -+•  jo  y^>2  __  4^^  _  4^2 


(13), 
(14). 


-  4^/,       4^2 
4(A+^) 

Problem   XVIII. —  To  determine  a  right-angled  triangle,  having 

given  the  hypothenuse  and  the  difference  of  two  lines  drawn  from  the 

two  acute  angles  to  the  centre  of  the  inscribed  circle. 
\ 

Let  ABC  be  a  right-angled  triangle,  right-angled  at  13,  and  AO,  OC, 

two  lines  drawn  from  the  acute  angles  to  the  centre  O  of  the  in- 
scribed circle. 

Let  AC  =  £;  AO  -  OC  =  d,  OC~*; 
then,  AO  =s  x  -f  d 

Produce  AO,  and  from  C  draw  CD  per- 
pendicular to  the  prolongation,  meeting  it 
at  D.  Then,  since  the  sum  of  the  angles 
BAC    and  ACB    is  equal    to   a   right-angle 

(Bk.  I,  Prop.  25,  Cor.  4),  and  since  the  lines  AO  and  CO  bisect 
these  angles  (Bk.  Ill,  Prob.  14),  OAC  -f  ACO  is  equal  to  a  haif 
a  right-angle. 

Since  the  outward  angle  COD  is  equal  to  the  sum  of  the  inward 
angles  (Bk.  I,  Prop.  25,  Cor.  6),  it  is  equal  to  half  a  right-angle; 
and  hence,  OCD  is  equal  to  half  a  right-angle,  and  hence  OD  and 
CD  are  equal.     Denote  either  by  z  ;  then, 


Then, 


x2  ±=  z1  -f-  z2 ;     and    z  =s  x  <y/\- 
AC2  sa  AD2  -f  CD2  :     that  is, 


V  =  (x  +  d  +  x  i/i?  +  (z  i/i)*; 
and  by  reduction,     (2  +  y^2)  x*+  d (2  +  -y/2)  x  =  h2-d*  ;     or, 
x2  -f-  dx 


:-2 


d* 


2+^/2 


PROBLEMS    FROM    LEG  END  RE. 


197 


hence, 


^-ptf/**-*-™" 


2+v=r 


--.**-> 


\/- 


-(2-V^)^2 


Let       OC  =  s  =  m,     AO  =  m  +  d,     CD  =  z  =  m-y/j, 
AB  =  y,     and     CB  =  «. 
Then,  since  the  triangles  ACD  and  AOG  are  similar, 


h  :  my/%  :  :  m  +  a  :  OG  =  r  =  —      h    ~> 


and 


h  :  w  4-  d  +  *»y  J  :  :  m  +  d  :  AG  — 
AB  =  y  =  AG  +  GB 


h 


(ra+eZ)  (m  +  d+m-T/2) 


=  6 ;     and 


u  =  -y/h2  -  b'K 

Problem  XIX. —  To  determine  a  triangle,  having  given  the  base,, 
the  perpendicular,  and  the  difference  of  the  two  other  sides. 

Let  ACB  be  a  triangle,  and  CD  perpendicular  to  the  base  AB. 

Let    AB  =  6,     CD  =  «,     AC  -  BC  =  d, 
BC  =  x,     AC  =  x+d,  and  DB  =  z, 
and  AD  =  b—z. 

Then, 


and 


AC=  Va2  +  (b  - zf  ; 
CB  r=  -y/a2  4-  zl ;     hence, 


^/a2  +  (6  _  2)2  _  -/a2  4-  z< 
Transposing  and  squaring,  we  have 


a*+(b-  z)2  =  d*  4-  2<2  ^/a2  4-  *2  4-  a2  +  *2  J 


198  APPENDIX, 

squaring  b  —  z}  in  the  first  member,  we  have 


a2  +  b*  —  2bz  +  z2  =  d2  +  2d  v^2  +  *2  +  a2  +  s2; 
whence,  by  reducing, 


(4»  -  <pj  -  262  =  2rf  y^2  +  *2 ; 
squaring  both  members,  we  have 

(62  _  tf2)2  _  4b  (£2  _  d2^  g  +  ±b2z*  _  4a2tf2  +  4^2  . 

transposing,  and  collecting  the  terms, 

4  (b2  -  d2)  .  z2  -  45  (62  -  d2)  .  ^  ss  4a2^  -  (b2  -  tf2)2 ;     or, 

4a2tf2  _  (£2  _  ^2)2 

f~-**-       4(^-lrf») ;     Whence' 


6        1     / 


4a2rf2  +  62d2  —  rf4 


b2-d2  ■ 

from  which  the  sides  AC  and  CB  are  easily  found. 

Problem  XX. — To  determine  a  triangle,  having  given  the  basef  the 
perpendicular,  and  the  rectangle  of  the  two  sides. 

Let  ACB  be  a  triangle,  and   CD  a  line 
drawn  perpendicular  to  the  base  AB. 
Let      AB  =  6,     CD  =  d,     AC  x  CB  =  q  ; 
AC  =  ar,     and     BC  s;  y ;     then, 
xyz=q      ...     (1). 
AD  =  V*2  -  d2,     and     DB=vy-d2;     hence, 


^2  _  d*  +  yy  _  ^2  =  6     ...     (2). 

Transposing,  ani  then  squaring  both  members  of  equation  (2),  we 
have 

x2  -  d2  =  b2  —  2b^/y2—  d2  4-  y2  —  d2 ; 


PROBLEMS  FROM  LEGENDRE.  199 

whence,         (x2  —  y2)  —  b2  a  —  2b  -y/y2  —  d2  ;     squaring  again, 
x*  -  2x2y2  +  y*-  2b2  (x2  -  y2)  +  b*  =  4%2  -  462tf2 ;  reducing, 
x*  -  2*2y2  +  y*  -  2b2(x2  +  y2)  =  -  b*  -  4b2(P     •     •     (3). 
Adding  4  times  the  square  of  each  member  of  (1), 

x*  +  2x2y2  +  y2  -  2b2  (x2  +  y2)  -  4q2  -  6*  —  4b2d2 ;     or, 
K?2  +  y2)2  —  %b2  {x2  +  y2)  =  4q2  —  b*  —  4b2d2. 
Regarding  x2  +  y2  as  a  single  unknown  quantity,  we  have 


x2  +  y2  =  b2  ±  -v/4?2  -  4  W     .     .     .     (4)  ; 
then,  by  adding  twice  each  member  of  equation  (1), 


x2  +  2.ry  +  x2  =  2q   +  b2  ±  -y/4?2  -  462d2 ;     or, 


*  +  y  =  y/2q    +  b2  ±  -x/4q2  -  462d2  : 
subtracting  from  equation  (4),  twice  equation  (1), 


m 


x2  —  2xy  +  x2  =  —  2q    +b2  ±  -y/4q2  —  4b2d2  ;     or, 

x  —  y  =  ±  -/—  2?   +  62  ±  ^/4^2  -  4«2  =  n  ; 

m  -\-  n  m  —  n 

hence,  x  ==  — -— ,     and    y  =  — - — • 

Problem  XXI. —  To  determine  a  triangle,  having  given  the  lengths 
cf  three  lines  drawn  from  the  three  angles  to  the  middle  of  the  opposite 
sides. 

Let  ACB  be  a  triangle,  and  AE,  BG,  CD,  three  lines  drawn  from 
the  vertices  to  the  middle  points  of  the  opposite  sides. 
Let     CD=a,     AE  =  b,     BG=c,     AC=*,      . 
BC=y,     and     AB=z:     then, 


x2  +  y2  =  2a2  +  ~  (Bk.  IV.  Prop.  14), 

lit 


Z2+Z*  =  2b2  +  ^;      y2+z2=2c2  +  ^  J 


200  APPENDIX. 

clearing  of  fractions  and  transposing,  we  have 
2x2  4  %V2  ~  22  =  4a2     •     •     •     (1) 
2x2  +  2z2  —  y2  =  4o2     •     •     .     (2) 
2y3  -f  2z2  -  x2  =  4c2     •     •     •     (3)  ; 
subtracting  equation  (2)  from  (1),  we  have 

3y2  -  3z2  =  4  (a2  -  b2) 
multiplying  equation  (3)  by  2,  and  adding  to  (2), 

3y2  +  6z2  =  8c2  +  462     .     .     (5)  ; 
subtracting  equation  (4)  from  (5), 

9s2  =  8c2  +  4o2  —  4«2  +  4o2 ;     hence, 


z=  ±1  y/2b2  +  2c2 


x  Vs  ±  }  V2a2  -f  262 


y  =  ±  §  -/2a2  +  2c2  —  b2. 

Cor.  Through  G  and  E,  the  middle  points  of  AC  and  BC,  draw 
GE ;  then  will  GE  be  parallel  to  the  base  AB  (Bk.  IV.  Prop.  16)  : 
and  since  AC  is  equal  to  twice  CG,  AD  will  be  equal  to  twice  GP, 
or  its  equal  PE. 

But  the  triangles  AOD  and  POE  are  similar ;  and,  since  AD  is 
equal  to  twice  PE,  AO  is  equal  to  twice  OE  ;  that  is, 

If  three  lines  be  drawn  from  the  vertices  of  the  three  angles  of  a 
triangle  to  the  middle  points  of  the  opposite  sides,  the  distance  from 
either  vertex  to  the  point  of  intersection,  will  be  two-thirds  of  the 
bisecting  line. 

Problem  XXII. — In  a  triangle,  having  given  the  three  sides,  to 
find  the  radius  of  the  inscribed  circle. 

Let  ABC  be  a  triangle,  CD  perpendicular  to  the  base,  O  the 
centre,  and  OP  the  radius  of  the  inscribed  circle. 


PROBLEMS  FROM  LEGENDRE. 


201 


Let  BC  =  a,    AB  =  c,    AC  =  b, 

OP  =  r,     AD  =  zt     CD  =  x. 

x2  =  b2-z2     .     .     .      (1), 
x2  =  a2  —  (c  —  z)2  ;     whence,  X 

b2  —  z2  =  a2  —  c2  -f-  2cz  —  z2 ;     therefore, 
52  4.  c2  -  a2 

e  =  — 27— ; 

combining   with  equation  (1), 

^f\b2c2  —  (b2  +  c2  -  a2)2 

*~  ge  ; 

area  ™  (u  -f-  i«  4"  tf)  4f  =  ^cx  j     whence, 
<•*  .,/4/>V  __  ^2  +  c2  —  a2)2 


D       B 


rt*f6-J-c  2(&-f-6  +  c) 

pROBf^.M  .VXTEL — To  de!enn:i>t  .1  tight-angled  triangle,  having 
given  the  sid*,  of  the  inscribed  tqw-.t  yid  the  radius  of  the  inscribed 
circle. 

Let  ABC  be  a  right-angled  tiiangV,  y*i*j»  «\  square  and  circle  both 
inscribed. 

Let       AB  =  x,      BC  =  y,     and      AC  •   z: 
denote  the  side  of  the  square  by  s,  and  tht  avl;i  * 
of  the  circle  by  r. 

Then,     x  +  y  -  z  =  2r  (Prob.  XIII)     i     (1) 

x2  +  y2  =  z2 (2)t    V 

and  x   +  y  =  *  +  -v/22  +  s2  (Prob.  XY)     •     •     <2> 

Combining  equations  (1)  and  (2),  and  transpose g, 


2  -f  2r  —  5  =  -yjz2  -f  *2  ;     squaring. 
z2  +  4r2  -f-  s2  +  4rz  —  2$z  —  4rs  =  z2  +  s2  t 


202  APPENDIX. 

.  2rs  —  2r2 

whence,  z  =    — t       .     .     .     (4) . 

combining  equations  (4)  and  (1), 

x  -f  y  =  o ;  =  mi     and  equation  (2)  gives 

2,  (2rs-2r2)2  • 

+  y        (2r  -  s)2      =       '    whence> 


ar  —  \  (m  +  ^/2n2  —  m2), 


y  =  1  (m  —  y^i2  —  m2). 

Problem  XXIV. — To  determine  a  right-angled  triangle,  having 
given  the  hypothenuse  and  radius  of  the  inscribed  circle. 

Let  ABC  be  a  right-angled  triangle,  and  O   the  centre  of  the 
inscribed  circle. 

Let        AC  =  h,       AB  =  x,       BC  =  y, 
and  r  =  the  radius  of  the  circle. 

x   +  y   =  A+2r(Prob.XIII)     .     (1) 
j£-^ts£ (2).      j£ 

Since  the  perimeter  is  equal  to  2A  -f-  2r,  and  since  four  times  the 
area  is  equal  to  the  perimeter  into  2r  (Prob.  XIII), 

2xy  —  4r2  -f  Mr     •     .     .     (3)  ; 
subtracting  equation  (3)  from  (2),  and  extracting  the  square  root, 


x  —  y  =  -y/A2  —  4r2  —  4rh     •     •     •     (4)  ; 
combining  (4)  and  (1), 


h  +  2r  +  -yA2  —  4r2  -4rh 
2  ' 


A  +  2r  —  -vA2  —  4r2  — 4rA 


PROBLEMS  FROM  LEGENDRE. 


203 


Problem  XXV. —  To  determine  a  triangle,  having  given  the  base, 
the  line  bisecting  the  vertical  angle,  and  the  diameter  of  the  circum- 
scribing circle. 

Let  ABC  he  a  triangle,  EF  the  diameter  of  the  circumscribing 
circle  perpendicular  to  AB,  and  CD  the  line  bisecting  the  vertical 
angle  :  this  line  will  pass  through  E,  the  middle  point  of  the  arc 
AEB. 


AB  =  b 
CD  =  a 

EF  s  d 
KO  =  x 


BC  =  y    I  DE  =  v 


Auxiliaries. 
EO  =  z;     OY=d-z 
DO  =  « 


z{d-z)  =  ^-     (Bk.  IV.  Prop.  28) ; 


wnence, 


d  ±  Jd2  -  b2 

Z  =  T-7Z =    l> 


By  drawing  CF,  we  have  two  similar  right-angled  triangles,  EOD 
and  EFC ;  hence, 

d    :     a  -f-  v     :  :     v  :     Z;     or, 

v2  +  av  =  dl ;  whence, 


—  a  ±  J  a2  +  4dl 

v  = £■ =  m 


also,  u  —  ^/v2  —  z2  =  -\/m2  —  I2  =  n. 

Then,         *y*  S a*  f  fj +  »)   (|-  -  »)     (Bk.  IV.  Prop.  31), 


and, 


5    ,  6 


204: 


APPENDIX. 


+  n 


whence,      x=yx 


and       y  =  x  x 


b  t 


substituting  these  values  in  the  preceding  equation,  and  reducing, 
1      /(4a2  +  b2  -  4n2)  (6  H-  2») 


6  —  2w 


y=4\/ 


1      /(4a2  4-  A2  -  4  m2)  (6  -  5>n) 


(b  +  2») 


PROBLEM  OF  THE  THREE  POINTS. 

CONSTRUCTION. 

From  a  station  P  there  can  be  seen  three  objects,  A,  B  and  C, 
whose  distances  from  each  other  are  known  ;  viz.,  AB  =  800  yards, 
AC  be  600  yards,  and  BC  =  400  yards :  there  are  also  measured 
the  horizontal  angles,  APC  =  33°  45'  and  BPC  =  22°  30';  it  is 
required  to  find  the  three  distances,  PA,  PC  and  PB. 

Lay  off  from  a  scale  of  equal  parts 
AB  =  800  yards ;  then,  with  the  known 
sides  AC  and  CD  construct  the  triangle 
ACB.  At  A  lay  off  an  angle  equal  to 
BPC  =  22°  30',  and  at  B  lay  off  an  angle 
equal  to  APC  —  33°  45',  and  through  the 
point  D,  where  these  lines  intersect,  and 
the  two  points  A  and  B,  describe  the  cir- 
cumference of  a  circle.     Join  C  and   D, 

and  produce  it  till  it  meets  the  circumference  at  P.  Then,  since  the 
angle  DAB  =  CPB,  standing  on  the  same  arc  DB,  and  the  angle 
ABD  =  APC,  standing  on  the  same  arc  AD,  it  follows  that  P  will 
be  the  position  of  the  third  point. 


PROBLEMS    FROM  LEGENDRE.  205 

CALCULATION. 

In  the  triangle  ABC,  we  have  given  the  three  sides ;  hence,  we 
can  find  the  angles  CAB  and  CBA  (Case  IV.  Trigonometry). 

In  the  triangle  ADB  we  know  the  base  AB  and  the  angles  at  the 
base ;  hence,  by  Case  I.,  Trigonometry,  we  can  find  the  sides  AD 
and  BD. 

From  the  found  angle  CAB,  subtract  the  known  angle  DAB 
=  22°  30',  and  we  have  the  angle  CAD. 

Then,  in  the  triangle  CAD,  we  know  two  sides  and  the  included 
angle,  and  hence,  can  find  the  angle  ACP  by  Case  III. 

Then,  in  the  triangle  ACP,  we  know  the  side  AC  ss  600  yards, 
and  the  two  angles  ACP  and  APC ;  hence,  we  can  find  PA  and  PC, 
after  which  we  can  easily  find  PP. 

PA=71(U93  yards;  PC=  1042.522  yards ;  PP= 934.291  yards. 


f»S^ 


THE 


[TJjriVBESITY; 


THZ     END. 


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